Suppose 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter. The final temperature of the water and copper calorimeter is 18.0C.
1) What was the initial common temperature of the water and copper? (Express your answer to three significant figures.)

Answers

Answer 1

The intital common temperature of copper and water is 9.5°C, under the condition that 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter.

Now to evaluate the initial common temperature of the water and copper calorimeter, we have to apply the formula
m1c1(Tk - Ti) + m2c2(Tk - Ti)
= mcopperccopper(Tk - Ti)

Here,
m1 = mass of water,
c1 =specific heat capacity of water,
m2 = mass of copper calorimeter,
c2 = specific heat capacity of copper calorimeter, mcopper = mass of copper block
ccopper =specific heat capacity of copper.

Here, this equation to evaluate Ti
Ti = (m1c1Tk + m2c2Tk - mcopperccopperTk - m1c1Ti - m2c2Ti) / (m1c1 + m2c2 - mcopperccopper)

Staging the given values into this equation
Ti = (-300.0 g)(4.18 J/g°C)(18.0°C) + (200.0 g)(0.385 J/g°C)(18.0°C) + (10.0 g)(0.385 J/g°C)(18.0°C) / [(300.0 g)(4.18 J/g°C) + (200.0 g)(0.385 J/g°C) - (10.0 g)(0.385 J/g°C)]
Ti = 9.5°C

Hence, the initial common temperature of the water and copper calorimeter was 9.5°C.
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Related Questions

Here are some data from a similar experiment, to determine the empirical formula of on oxide of tin.
Calculate the empirical formula according to these data.
Mass of crucible, cover, and tin sample 21.76 g
Mass of empty crucible with cover 19.66 g
Mass of crucible and cover and sample,
after prolonged heating gives constant weight 22.29 g

Answers

The information given can be used to construct the empirical formula for a tin oxide. We must first determine the mass of tin in the sample. This may be achieved by deducting the mass of the crucible, cover, and sample (21.76 g) from the mass of the empty crucible and cover (19.66 g).

This gives us a mass of 2.10 g of tin in the sample. The mass of oxygen in the sample must then be determined. To achieve this, we must deduct the mass of the crucible, cover, and sample (21.76 g) from the mass of the same components (22.29 g) prior to protracted heating. This provides us with an oxygen mass of 0.53 g.

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1. To operate a batch reactor for converting A into R. This is a liquid phase reaction with the stoichiometry A → R. CA,(mol/l) 0.1 0.2 0.3 0.4 0.2 0.6 0.7 0.8 1.0 1.3 2.0 -rA,(mol/l min) 0.1 0.3 0.5 0.6 0.5 0.25 0.10 0.06 0.05 0.045 0.042 For the above data determine the order of reaction and rate constant.​

Answers

The reaction is second order with a rate constant of 0.043 mol/l min.

How to explain the reaction

For CA = 0.1 mol/l, -rA = 0.1 mol/l min

For CA = 0.2 mol/l, -rA = 0.3 mol/l min

For CA = 0.3 mol/l, -rA = 0.5 mol/l min

For CA = 0.4 mol/l, -rA = 0.6 mol/l min

The slope of this line is equal to the order of the reaction (n), and the y-intercept is ln(k).

Slope = (0.6931 - (-2.3026)) / (0.3010 - (-0.9163)) = 1.929

ln(k) = -2.3026 + 1.929 * (-0.3010)

ln(k) = -3.1504

k = e^(-3.1504) = 0.043 mol/l min

The reaction is second order with a rate constant of 0.043 mol/l min.

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Why is leaf called the kitchen of the plant?

Answers

Answer:

Leaves of the plant contain a green colour pigment called chlorophyll. This pigment is responsible for trapping of sunlight which is essential for photosynthesis. This energy is used to synthesise food from carbon dioxide and water. Hence, leaves are called kitchen of the plant.

Plants rely on their leaves to produce food through a process called photosynthesis. This involves converting light energy into organic compounds, like sugars, using chloroplasts that contain the pigment chlorophyll. By combining carbon dioxide and water with light energy, plants create glucose, which serves as an energy source and building material. Along with stomata, which help regulate gas exchange with the environment, the leaf acts as the plant's primary kitchen for food production.

Which of these are part of the
Earth's lithosphere?

O clouds
O glaciers
O mountains
O water vapor

Answers

Mountains is the correct answer

Macmillan Learning Determine the formal charge on each atom in the structure. H H-B-H H What is the overall charge on the structure? -2 +1 Answer Bank +2 +3 -3 -4 +4 0 ​

Answers

The overall charge on the structure is negative one (-1).

The central boron atom in the structure is bonded to two hydrogen atoms. Boron has three valence electrons, and it has formed only two bonds, so it has a formal charge of +1.

Each of the hydrogen atoms has one valence electron, and each is bonded to the boron atom, so each hydrogen atom has a formal charge of -1. The sum of the formal charges in the structure is equal to the charge of the ion, which is -2. Adding up the formal charges of the atoms, we get:

B: +1

H: -1 (two times)

Overall charge = sum of formal charges = +1 - 1 - 1 = -1

Therefore, the overall charge on the structure is negative one (-1).

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diffrences in water temperature in the ocean create movement because-

Answers

Diffrences in water temperature in the ocean create movement because bodies of water at different temperatures have different densities.

How can the differences  be explained?

Water that is colder is generally denser than water that is warmer, so when a body of water with colder, denser water is next to a body of water with warmer, less dense water, a density gradient is established. This gradient creates a difference in pressure between the two bodies of water, with the colder, denser water being at a higher pressure than the warmer, less dense water.

This difference in pressure creates a force that drives the movement of water from the denser, colder region to the less dense, warmer region. This movement of water is known as convection, and it can occur both vertically and horizontally in the ocean. Vertical convection occurs when differences in temperature cause water to rise or sink, while horizontal convection occurs when water moves laterally due to differences in temperature.

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missing options:

1. as water heats up, the atoms of water more faster.

2. warm water is pulled more by gravity than cold water.

3. warm and cold water mix and reach the same temperature.

4. bodies of water at different temperatures have different densities.

What is the difference between practical work inside a laboratory and outside a laboratory

Answers

Answer:

The main difference between practical work inside and outside a laboratory is that the practical work inside the lab includes good equipment and chemicals which are very advanced and the practical outside a laboratory is more about the safety of life.

Explanation:

Practicals are set up at stations with lab equipment and chemicals, where students can learn, and researchers can experiment and find different new things.

Thus, the practical work inside the lab includes lab equipment and chemicals, and the practical outside a laboratory is more about conserving nature.  

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Convert the following number
into correct scientific notation.
0.0602 x 10^25

[ ? ] × 10 [ ? ]

Answers

The number is converted to 60. 2 × 10²²

What are index forms?

Index forms are simply described as mathematical forms that are used in the representation of numbers that are too small or too large in more convenient forms.

These index forms are also referred to as scientific notation or standard forms.

Some rules of index forms are;

Add the exponents when multiplying forms of the same basesSubtract the exponents when dividing forms of the same bases

From the information given, we have that;

0. 0602 × 10 ²⁵

Subtract three from the exponent value and move three spaces right, we have;

60. 2 × 10²⁵⁻³

60. 2 × 10²²

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Chemistry. . . Reaction: AB₂C (g) → B₂ (g) + AC (g), find the value of K
At equilibrium [AB₂C]=0.0168 M, [B₂]= 0.007 M, and [AC] = 0.0118 M

Answers

The value of K at equilibrium, for the reaction is 0.0049

How do i determine the value of K at equilibrium?

First, we shall list out the given parameters from the question. This is shown below:

AB₂C (g) ⇌ B₂(g) + AC(g) Concentration of AB₂C, [AB₂C] = 0.0168 MConcentration of B₂, [B₂]= 0.007 MConcentration of AC, [AC] = 0.0118 MEquilibrium constant (K) =?

Equilibrium constant is defined as:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

m is the coefficient of productsn is the coefficient of reactants

With the above formula, we can obtain the equilibrium constant, K as follow:

Equilibrium constant, K = [B₂][AC] / [AB₂C]

K = (0.007 × 0.0118) / 0.0168

K = 0.0049

Thus, the equilibrium constant, K for the reaction is 0.0049

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A gas‑filled weather balloon has a volume of 56.0 L
at ground level, where the pressure is 761 mmHg
and the temperature is 23.1 ∘C.
After being released, the balloon rises to an altitude where the temperature is −6.97 ∘C
and the pressure is 0.0772 atm.
What is the weather balloon's volume at the higher altitude?

Answers

To determine the weather balloon's volume at the higher altitude, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

P1V1/T1 = P2V2/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

Converting the initial conditions to SI units:

P1 = 761 mmHg = 101.325 kPa
V1 = 56.0 L
T1 = 23.1 + 273.15 = 296.25 K

Converting the final conditions to SI units:

P2 = 0.0772 atm * 101.325 kPa/atm = 7.84 kPa
T2 = -6.97 + 273.15 = 266.18 K

Solving for V2:

V2 = V1 * P1 * T2 / (P2 * T1)
V2 = 56.0 * 101.325 * 266.18 / (7.84 * 296.25)
V2 = 122.7 L

Therefore, the weather balloon's volume at the higher altitude is 122.7 L.

Predict which of the following reactions has a positive change in entropy.
I. 2N2(g) + O2(g) → 2N2O(g)
II. CaCO3(s) → CaO(s) + CO2(g)
III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)​

Answers

Answer:

Explanation:

The change in entropy of a system can be determined by comparing the entropy of the reactants to the entropy of the products. The reaction that leads to an increase in the number of moles of gas or particles will generally have a positive change in entropy.

I. 2N2(g) + O2(g) → 2N2O(g)

The reactants have 3 moles of gas, while the product also has 3 moles of gas. Therefore, there is no change in the number of moles of gas, and the change in entropy is likely to be small.

II. CaCO3(s) → CaO(s) + CO2(g)

The reactant is a solid, while the products are a solid and a gas. The formation of a gas from a solid leads to an increase in the number of moles of particles, and therefore an increase in entropy.

III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

The reactants consist of a solid and a liquid, while the products consist of an aqueous solution and a gas. The formation of a gas leads to an increase in the number of moles of particles, and therefore an increase in entropy.

Therefore, reactions II and III have a positive change in entropyentropy

Gaseous butane (CH3(CH2)2CH3) will react with gaseous oxygen (02) to produce carbon dioxide (CO2) and gaseous water (H2O). Suppose 34.g of butane s mixed with 200. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answers

The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.

Determining the maximum mass of water produced

The balanced chemical equation for the reaction between butane and oxygen is:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.

moles of butane = 34. g / 58.12 g/mol = 0.585 mol

moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol

Determining the limiting reactant.

butane : oxygen = 0.585 mol : 6.25 mol

= 0.0936 : 1.00

stoichiometric ratio = 1 : 13/2

= 0.7692 : 1.00

Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.

The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).

moles of water = 5/13 * 6.25 mol

= 2.403 mol

Finally, we can convert the moles of water to grams:

mass of water = 2.403 mol * 18.015 g/mol

= 43.3 g

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What is the molar mass of a compound if a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr? The temperature in Celsius is known to two significant figures.

Answers

If a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr,  the molar mass of the compound is 24.8 g/mol.

To calculate the molar mass of the compound, we first need to calculate the number of moles present in the gaseous sample using the ideal gas law:

PV = nRT

Where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.

Converting the given pressure of 615 torr to atm:
615 torr = 0.811 atm
Converting the given temperature of 30°C to Kelvin:
30°C + 273.15 = 303.15 K

Rounding off to two significant figures, we get:
P = 0.81 atm
T = 303 K

Now, rearranging the ideal gas law equation to solve for n:
n = PV/RT

Substituting the given values:

n = (0.978 g/L) x (1 L) / (0.081 atm x 0.0821 L atm/mol K x 303 K)

n = 0.0394 mol

Next, we can calculate the molar mass of the compound using the formula:

molar mass = mass / mole
molar mass = (0.978 g/L) x (1 L) / 0.0394 mol
molar mass = 24.8 g/mol
Therefore, 24.8 g/mol is the molar mass of the compound.

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Describe the technique for washing a precipitate. Place the steps in the correct order.
A. add deionized water
B. mix solutions
C. decant
D. centrifuge

Answers

Answer: A.B.D.C Explanation: Took it

Chromium, Cr, has the following isotopic masses and fractional abundances:
Mass Number Isotopic Mass (amu) Fractional Abundance
50 49.9461 0.0435
52 51.9405 0.8379
53 52.9407 0.0950
54 53.9389 0.0236
What is the atomic mass of chromium

Answers

The average mass of chromium is 52.1. Isotopic mass is defined as the average mass of all the isotopes of a specific element.

The average atomic mass of an element is referred to as the sum of the masses of its isotopes, each multiplied by its natural abundance which can be also explained as the decimal associated with the percent of atoms of that element that are of a given isotope. Average atomic mass is equal to f1M1 + f2M2 and so on. Hydrogen, chromium, lithium, cobalt, oxygen, boron, plutonium, and carbon are some examples.

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The 500 cm³ of a pas enclosed in a container under a pressure of 580 mm of Hg. If the volume is reduced to 300 cm³ what will be the pressure then? ​

Answers

Answer:

The answer is 966.67 mm of Hg.

Explanation:

To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:

P1V1 = P2V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Using the given values:

P1 = 580 mmHg

V1 = 500 cm³

V2 = 300 cm³

We can solve for P2:

P1V1 = P2V2

580 mmHg x 500 cm³ = P2 x 300 cm³

290,000 mmHg·cm³ = P2 x 300 cm³

P2 = 290,000 mmHg·cm³ / 300 cm³

P2 = 966.67 mmHg (rounded to the nearest hundredth)

Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.

What is the molar volume of CO2 at 39 C and 652 torr?

Answers

The molar volume of a gas can be calculated using the ideal gas law:

PV = nRT

where P is the pressure of the gas in atmospheres (atm), V is the volume of the gas in liters (L), n is the number of moles of gas, R is the ideal gas constant (0.08206 L·atm/mol·K), and T is the temperature of the gas in Kelvin (K).

To solve for the molar volume of CO2 at 39°C (312 K) and 652 torr (0.859 atm), we can rearrange the ideal gas law as follows:

V = (nRT) / P

First, we need to calculate the number of moles of CO2. We can use the following equation, which relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

Solving for n, we get:

n = (PV) / (RT)

Substituting the given values, we get:

n = (0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)

Now we can substitute this expression for n into the equation for the molar volume:

V = (nRT) / P

V = [(0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)] * (0.08206 L·atm/mol·K * 312 K) / (0.859 atm)

Simplifying, we get:

V = 24.45 L/mol

Therefore, the molar volume of CO2 at 39°C and 652 torr is 24.45 L/mol.

1. Which of the following stars has a temperature of approximately 9000 K and luminosity about to
20 times greater than the Surfs luminos

a Sirius
b. Procyon
c. Figel
d. Polaris


2. Which of the following types of stars is considered part of the main sequera

a Supergants
b. Red giants
c. Red dwarts
d. White dwarfs


3. Which of the following stars is cooler than the
Surf

a. Procyon B
b. Pigel
C. Barnard's Star
d. Sirius


4. The Sun is classified with which of the following types of stars?

a. Supergiants
b. Red giants
c. Main sequence
d. White dwars


5. Which of the forces listed below is most responsible for the formation of start?

a. Gravity
b. Magnetism
c. Bectromagnetism
d. Light


6. Which star has a higher luminosity and a lower temperature than the Sun?

a. Pigel
b. Barnard's Star
c. Alpha Centauri
d. Aldebaran


7. Compared to the temperature and luminosity of the star Polars, the star Srus is

a. hotter and more luminous
b. hotter and less luminous
c. cooler and more luminous cooler and less luminous

Answers

1. The star that has a temperature of approximately 9000 K and luminosity about 20 times greater than the Sun’s luminosity is Vega.

2. The type of star that is considered part of the main sequence is red dwarfs.

3. The star that is cooler than the Sun is Barnard’s Star.

4. The Sun is classified as a main sequence star.

5. The force most responsible for the formation of stars is gravity.

6. The star that has a higher luminosity and a lower temperature than the Sun is Aldebaran.

7. Compared to the temperature and luminosity of the star Polaris, the star Sirius is hotter and more luminous.

A solution contains 0.0400 M Ca2+ and 0.0990 M Ag+. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first?

Ca3(PO4)2
Ag3PO4
Na3PO4
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answers

15.66% of the first cation is still in solution as the second cation is just beginning to precipitate.

What is phosphate used for?

One of the three main nutrients that are most frequently used in fertilisers is phosphorous, which is obtained from processing phosphate rock (the other two are nitrogen and potassium).

You can also make phosphoric acid into phosphoric acids, which are utilised in everything from food and skincare to animal feed and electronics. Over the course of millions of years, organic matter accumulates to form the sedimentary rock known as phosphate.

When [tex]Na_{3}Po_{4}[/tex] added to the solution of Ca and Mg, [tex]Ca_{3}(Po_{4})_{2}[/tex] and [tex]Ag_{3}Po_{4}[/tex] are formed.

Ksp of [tex]Ca_{3}(Po_{4})_{2} = 2.07*10^{-33}[/tex]

Ksp of [tex]Ag_{3}Po_{4} = 0.09*10^{-17}[/tex]

Concentration of [tex][Ca^{2+}][/tex] = 0.040 M

Concentration of [tex][Ag^{+}][/tex] = 0.0990 M

[tex]Ag_{3} Po_{4} - > 3Ag^{+} + Po_{4}^{3-}[/tex]

Ksp = [tex][Ag^{+}]^{3} [Po4^{3-}][/tex]

[tex]0.09*10^{-17} = (0.099)^{3} [Po_{3-}][/tex]

[tex][Po_{3-}] = 9.16*10^{-14}M[/tex]

[tex]Ksp = [Ca^{2+}]^{3} [Po_{3-}][/tex]

[tex]2.07*10^{-33} = (0.040)^{3} [Po_{4}^{3-}]^{2}[/tex]

[tex][Po_{4}^{3-}] = 5.68*10^{-15} M[/tex]

[tex][Po_{4}^{3-}][/tex] is smaller in [tex]Ca_{3}(Po_{4})_{2}[/tex]

[tex]Ca_{3}(Po_{4})_{2}[/tex] will start precipitating first

[tex]Ksp = [Ca^{2+}]^{3} [Po_{4}^{3-}]^{2}[/tex]

[tex]2.07*10^{-33} = [Ca^{2+}]^{3} (9.16*10^{-14})^{2}[/tex]

[tex][Ca^{2+}] = 6.27*10^{-3} M[/tex]

[tex]\%\ of\ Ca^{2+}[/tex] remaining [tex]= 6.27*10^{-3}/0.040 * 100[/tex]

= 15.66 %

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A specific organic reaction is described by the energy diagram drawn below. Using this
energy diagram, identify which product will form first and which product will be the
major product if given enough time?

Answers

The product that is formed first is product B and the product that will be the major product if given enough time would also be product B.

In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products.

The activation energy of the reaction can be shown on a diagram as the energy between the reactants that the transition state.

The product with lesser activation energy is the product that is formed first and the major product is decided by the stability of the product which depends on the energy. Lesser is the energy of the product, greater is its stability.

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Aqueous sulfuric acid (H₂SO₂) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO) and liquid water (H₂O). What is the
theoretical yield of sodium sulfate formed from the reaction of 4.9 g of sulfuric acid and 5.0 g of sodium hydroxide?
Round your answer to 2 significant figures.

Answers

The theoretical yield of sodium sulfate, Na₂SO₄, formed from the reaction of 4.9 g of sulfuric acid, H₂SO₄ and 5.0 g of sodium hydroxide, NaOH is 7.1 g

How do i determine the theoretical yield?

First, we shall determine the limiting reactant. This is shown below:

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Molar mass of H₂SO₄ = 98 g/molMass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g Molar mass of NaOH = 40 g/molMass of NaOH from the balanced equation = 2 × 40 = 80 g

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH

Therefore,

4.9 g of H₂SO₄ will react with = (4.9 × 80) / 98 = 4 g of NaOH

From the above calculation, we can see that only 4 g of NaOH out of 5 g is needed to react with 4.9 g H₂SO₄.

Thus, the limiting reactant is H₂SO₄

Finally, we shall determine theoretical yield of sodium sulfate, Na₂SO₄ formed. Details below:

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Molar mass of H₂SO₄ = 98 g/molMass of H₂SO₄ from the balanced equation = 1 × 98 = 98 gMolar mass of Na₂SO₄ = 142 g/molMass of Na₂SO₄ from the balanced equation = 1 × 142 = 142 g

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄

Therefore,

4.9 g of H₂SO₄ will react to produce = (4.9 × 142) / 98 = 7.1 g of Na₂SO₄

Thus, the theoretical yield of sodium sulfate, Na₂SO₄ formed is 7.1 g

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Question 6 (5 points)
Label each situation as metals, nonmetals, or metalloids. Label each with numbers.

Usually conducts electricity
& heat well

at room temperature these
are gases or liquids

Will lose valance electrons
to form compounds.

can be used as
semiconductors

Will gain valance electrons
to form compounds.

1. a metal
2. a nonmetal
3. a metalloid

Answers

Answer:

Explanation:

Here are the labels for each situation:

1.Usually conducts electricity & heat well - Metal (1)

2.At room temperature these are gases or liquids - Nonmetal (2)

3.Will lose valance electrons to form compounds - Metal (1)

4.Can be used as semiconductors - Metalloid (3)

5.Will gain valance electrons to form compounds - Nonmetal (2)

CHEMISTRY chemistry Table balance A+B→C Table2

Answers

Answer:

zn + Hcl cual es su rreaccion

using the equation PCI5(g) PCI3(g) + CI2(g), if CI2 is added, what way will the euilibeium shift

Answers

When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

Thus, The adjustments required to reach equilibrium might not be as obvious if we have a mixture of reactants and products that have not yet reached equilibrium.

In this situation, we can compare the Q and K values for the system to forecast changes.

By adding or withdrawing one or more of the reactants or products, an equilibrium chemical system can be momentarily moved out of equilibrium. After that, additional adjustments are made to the reactant and product concentrations in order to bring the system back to equilibrium.

Thus, When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

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What is the equilibrium constant, K? 3 A(g) + 3 B(g) <-> 5 C(g) + 2 D(g)

Answers

The equilibrium constant is written as;

Keq = [tex][D]^2 [C]^5/[A] [B]^3[/tex]

What is the equilibrium constant?

The equilibrium constant's value is influenced by the reaction's chemical make-up and temperature.

The product of the product concentrations, each raised to the power of their stoichiometric coefficient, divided by the product of the reactant concentrations, each raised to the power of their stoichiometric coefficient, is known as the equilibrium constant.

The equilibrium constant is Keq = [tex][D]^2 [C]^5/[A] [B]^3.[/tex]

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What happens to a buffered solution when a small amount of base is added?
O The solution quickly becomes neutral.
O The solutions resists changes in pH.
O The solution slowly becomes acidic.
O The solution quickly becomes basic.

Answers

Answer:

solution resists changes in pH

Explanation:

the inherent property of buffers is to resist change to ph even when acids and bases are added. when the base is added, it is quickly neutralized by the conjugate acid, so the ph won't change.

Answer:

B: The solutions resists changes in pH.

Explanation:

Buffer reactions maintain stable pH of solutions.

Write the cations and anions present in CrO2

Answers

The chemical molecule CrO2 is also known as chromium(IV) oxide or chromic acid. It has the molecular formula CrO2 and is an inorganic substance.

In the solid state, CrO2 exists as a solid with a layered structure, and it is considered a cationic compound. The cation present in CrO2 is chromium(IV) ion, denoted as Cr4+.

On the other hand, the anion present in CrO2 is oxide ion, denoted as O2-. The oxidation state of oxygen in this compound is -2.

So, the cations present in CrO2 are Cr 4+ ions, and the anions present are O2 -2 ions.

In CrO2, the cation present is Chromium (Cr) with a charge of +4, and the anion present is Oxygen (O) with a charge of -2.

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Chemistry what is the reaction Rate TABLE

Answers

The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.

How to determine rate constant?

The rate of the reaction P → E + Z can be expressed as:

Rate = - d[P]/dt = d[E]/dt = d[Z]/dt

where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.

Use the experimental data to determine the rate constant and the order of the reaction.

Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:

rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s

rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s

rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s

The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.

Now using any of the three trials to determine the value of the rate constant k, trial 1:

Rate1 = k[P]1

k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹

Therefore, the rate law for the reaction P → E + Z is:

Rate = k[P]

where k = 5.27 E-2 s⁻¹ is the rate constant.

Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:

Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s

Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s

These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.

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If the reaction A (aq) + B (aq) C(aq) has a Ka value equal to 4.26 x 10-6, what is the G value at 25 °C if the concentrations are as follows:

[A] = 1.50 M
[B] = 1.00 M
[C] = 5.00 x 10-5 M

Answers

The Gibbs free energy change for the given reaction at 25°C and the given concentrations is -25.5 kJ/mol

The Gibbs free energy change (∆G) of a reaction can be calculated using the equation:

∆G = -RT ln(K)

Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, and K is the equilibrium constant.

The equilibrium constant (K) can be calculated from the acid dissociation constant (Ka) as:

K = [C] ÷ ([A] × [B])

Substituting the given values, we get:

K = (5.00 x 10⁻⁵) ÷ (1.50 x 1.00) = 3.33 x 10⁻⁵

Therefore,

∆G = - (8.314 J/molK) × (298 K) × ln(3.33 x 10⁻⁵)

= 25.5 kJ/mol

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Please help thanks so much!!!!!!!!!!!!

Answers

The total mass of products obtained when 130 g of zinc react completely with HCl is 274 g (3rd option)

How do i determine the total mass of products obtained?

First, we shall determine the mass of each product obtained. Details below:

For ZnCl₂

2HCl + Zn -> ZnCl₂ + H₂

Molar mass of Zn = 65 g/molMass of Zn from the balanced equation = 1 × 65 = 65 g Molar mass of ZnCl₂ = 135 g/molMass of ZnCl₂ from the balanced equation = 1 × 135 = 135 g

From the balanced equation above,

65 g of Zn reacted to produce 135 g of ZnCl₂

Therefore,

130 g of Zn will react to produce = (130 × 135) / 65 = 270 g of ZnCl₂

Thus, the mass of ZnCl₂ obtained is 270 g

For H₂

2HCl + Zn -> ZnCl₂ + H₂

Molar mass of Zn = 65 g/molMass of Zn from the balanced equation = 1 × 65 = 65 g Molar mass of H₂ = 2 g/molMass of H₂ from the balanced equation = 1 × 2 = 2 g

From the balanced equation above,

65 g of Zn reacted to produce 2 g of H₂

Therefore,

130 g of Zn will react to produce = (130 × 2) / 65 = 4 g of H₂

Thus, the mass of H₂ obtained is 4 g

Finally, we shall determine the total mass of the product produced. Details below:

Mass of ZnCl₂ = 270 gMass of H₂ = 4 gTotal mass of product =?

Total mass of product = mass of ZnCl₂ + mass of H₂

Total mass of product = 270 + 4

Total mass of product = 274 g (3rd option)

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