A sample of gas initially at 1. 4 atm and occupies 720 ml whats the final pressure in atm when the volume changes to 820 mL?

The solubility of the gas increases to 0.780 g/L when the pressure is increased to 131 kPa.

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Thus, we can use the following equation to calculate the new solubility:

S₂ ÷ S₁ = P₂ ÷ P₁

where S₁ is the initial solubility, S₂ is the new solubility, P₁ is the initial pressure, and P₂ is the new pressure.

Plugging in the given values, we have:

S₂ ÷ 0.650 g/L = 131 kPa ÷ 109 kPa

Solving for S₂, we get:

S₂ = (0.650 g/L) × (131 kPa ÷ 109 kPa)

S₂ = 0.780 g/L

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The volume of 53.5 g of O₂ at 30.1°C and 110.0 kPa is 1 m³ approximately

According to Charles' Law, while pressure is maintained constant, the volume of a given amount of gas varies in direct proportion to the absolute temperature of the gas. The Kelvin scale is used to measure temperature to determine the absolute temperature.

To find the volume of a gas, we can use the Ideal Gas Law:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the universal gas constant, and T is the temperature of the gas in Kelvin.

First, we need to convert the given temperature of 30.1°C to Kelvin:

T = 30.1°C + 273.15 = 303.25 K

Next, we need to determine the number of moles of O₂ present. We can use the molar mass of O₂ to convert from grams to moles:

molar mass of O₂ = 32.00 g/mol

moles of O₂ = 53.5 g / 32.00 g/mol = 1.671875 mol

Now we can rearrange the Ideal Gas Law to solve for V:

V = nRT / P

V = 1.671875 × 8.3145 × 303.25 /110 k × 1000 Pa / kPa

V = 0.062878 m³

Finally, we round the answer to the nearest tenth: (rounded to one decimal place) V = 1 m³

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The molarity of the solution containing 15.0 g of NaOH in 115.0 mL of H₂O is 3.26 M.

To calculate the molarity of the solution, we first need to convert the mass of NaOH and the volume of water to moles and liters, respectively.

First, we need to find the number of moles of NaOH in 15.0 g. The molar mass of NaOH is 40.00 g/mol, so:

15.0 g NaOH x (1 mol NaOH/40.00 g NaOH) = 0.375 mol NaOH

Next, we need to convert the volume of water from milliliters to liters:

115.0 mL H₂O x (1 L/1000 mL) = 0.115 L H₂O

Now we can calculate the molarity of the solution:

Molarity = moles of solute/liters of solution

Molarity = 0.375 mol NaOH / 0.115 L H₂O

Molarity = 3.26 M

Therefore, the molarity of the solution is 3.26 M.

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Answer: 0.942 moles of NaCl

Explanation:

for every 2 moles of Na3PO4 that react, 6 moles of NaCl form

therefore, to find how many moles of NaCl for we use this formula:

0.314 moles Na3PO4 * (6/2) = 0.942 moles of NaCl

A. The volume (in L) is 12.80 L

B. The mole is 0.035 mole

C. The temperature is 407.57 °C

D. The pressure is 126.98 atm

The volume can be obtained as follow:

PV = nRT

5.44 × V = 2 × 0.0821 × 424

Divide both sides by 5.44

V = (2 × 0.0821 × 424) / 5.44

Volume (V) = 12.80 L

The number of mole can be obtained as follow:

PV = nRT

0.250 × 1.80 = n × 0.0821 × 155

Divide both sides by (0.0821 × 155)

n = (0.250 × 1.80) / (0.0821 × 155)

Number of mole (n) = 0.035 mole

The temperature can be obtained as follow:

PV = nRT

4.47 × 26 = 2.08 × 0.0821 × T

Divide both sides by (2.08 × 0.0821)

T = (4.47 × 26) / (2.08 × 0.0821)

T = 680.57 K

Subtract 273 to obtain answer in °C

T = 680.57 - 273 K

Temperature (T) = 407.57 °C

The pressure can be obtained as follow:

PV = nRT

P × 2.25 = 10 × 0.0821 × 348

Divide both sides by 2.25

P = (10 × 0.0821 × 348) / 2.25

Pressure (P) = 126.98 atm

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A) The empirical formula of acid X is CH2O since it contains 31.6% carbon, 5.3% hydrogen, and 63.1% oxygen, b) the relative molecular mass of acid X is 34.2 g mol⁻¹.

An empirical formula is a chemical formula that indicates the simplest, whole number ratio of atoms in a molecule. It shows the types of atoms and the number of each type of atom that make up a single molecule of a compound.

a. The empirical formula of acid X is CH2O since it contains 31.6% carbon, 5.3% hydrogen, and 63.1% oxygen.

b. The number of moles of acid X in 11.4 g of the solution is 11.4/M, where M is the relative molecular mass of acid X. The number of moles of NaOH required to react with this amount of acid X is 0.100 mol dm⁻³ × 30 cm² = 0.03 mol. Thus, the mole ratio of acid X to NaOH is 11.4/M : 0.03, or M : 0.03 × 11.4/M. This can be rearranged to give M = 0.03 × 11.4/M, or M = 34.2 g mol⁻¹. Therefore, the relative molecular mass of acid X is 34.2 g mol⁻¹.

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The statements that correctly describe the strength of an acid or base are:

An acid is a chemical that donates hydrogen ions, whose addition to an existing solution results in increased acidity.

According to the conventional definition of acids, they are compounds which discharge positively charged hydrogen ions when mixed with water. Acids have a sour flavor and possess pH levels below 7.

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We need to dissolve 0.0405 mg mass of LiOH in 300.0 mL of water to get a solution with a pH of 11.25.

To find the mass of LiOH needed to make the solution, we need to first calculate the concentration of hydroxide ions in the solution using the pH value. Since pH = 11.25, the [OH⁻] concentration can be found by taking the negative logarithm of 11.25 and converting it to the concentration scale.

[tex][OH^-] = 10^{-11.25} = 5.62 \times 10^{-12} \, \text{M}[/tex]

Since LiOH is a strong base, it will dissociate completely in water, so the amount of LiOH needed can be calculated using the stoichiometry of the balanced equation:

LiOH + H₂O → Li⁺ + OH⁻ + H₂O

Thus, 1 mole of LiOH produces 1 mole of OH⁻. To achieve a concentration of 5.62 x 10⁻¹²M, we need 5.62 x 10⁻¹² moles of LiOH per mL of solution. Therefore, for 300.0 mL of solution, the number of moles of LiOH needed is:

[tex]\[5.62 \times 10^{-12} \, \text{mol/mL} \times 300.0 \, \text{mL} = 1.69 \times 10^{-9} \, \text{mol}\][/tex]

The molar mass of LiOH is 23.95 g/mol, so the mass of LiOH needed is:

1.69 x 10⁻⁹ mol x 23.95 g/mol = 4.05 x 10⁻⁸ g or 0.0405 mg (to 4 significant figures).

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2.50 litres of a 10.0 M solution require the preparation of 25.0 moles of ethylene glycol.

Excellent antifreeze, anti-boil, and anti-corrosive qualities are produced when antifreeze and water are mixed in a 50/50 ratio. The proportion of conventional ethylene glycol to water in severely cold conditions can reach 70% antifreeze, 30% water. The maximum antifreeze to water ratio when using DEX-COOL® is 60/40.

moles = concentration (M) x volume (L)

Given that the desired concentration is 10.0 M and the volume needed is 2.50 L, the setup for calculating the total number of moles of ethylene glycol can be written as:

moles = 10.0 M x 2.50 L

moles = 25.0 mol

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At 280 K and 1.50 atm, the mass of NaCl required to react with F₂ is 115.83 g; at STP, the mass of NaCl required to react with F₂ is 78.39 g.

Using the ideal gas equation, we will first determine the number of moles in F2:

Volume (V) = 15 L

Temperature (T) = 280 K

Pressure (P) = 1.5 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?  

                                F₂ + 2NaCl → Cl₂ + 2NaF

From the balanced equation above,

1 mole of F₂ reacted with 2 moles of NaCl.

0.98 mole of F₂ will react with = 0.98 × 2

                                          = 1.96 moles of NaCl

Mole of NaCl = 1.96 moles

Molar mass of NaCl = 58.5 g/mol

Mass of NaCl =?

At standard temperature and pressure (STP),

22.4 L = 1 mole of F₂

15 L = 15 / 22.4

15 L = 0.67 mole of F₂

                            F₂ + 2NaCl → Cl₂ + 2NaF

From the balanced equation above,

1 mole of F₂ reacted with 2 moles of NaCl.

0.67 mole of F₂ will react with = 0.67 × 2 = 1.34 moles of NaCl

Mole of NaCl = 1.34 moles

Molar mass of NaCl = 58.5 g/mol

Mass of NaCl =?

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Karl's leg cramp is unlikely to be caused by lactic acid, and the chemical equation for the process he is thinking of is C₆H₁₂O₆ + 2 ATP → 2 C₃H₃O₃⁻ + 2 NADH, option B is correct.

Karl's assumption that lactic acid is responsible for his leg cramp is a common misconception. In reality, lactic acid is a byproduct of anaerobic respiration, which occurs when there is not enough oxygen available to support aerobic respiration.

The process of glycolysis, which is the breakdown of glucose to pyruvate with the help of ATP. This process occurs in the cytoplasm of cells and is the first step in cellular respiration. The two pyruvate molecules produced by glycolysis can then be further broken down in the mitochondria to produce ATP through aerobic respiration, option B is correct.

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The complete question is:

Karl is at the gym exercising. After a while on the treadmill, he gets a cramp in his legs. Karl blames lactic acid building up in his muscles. What is the chemical equation for this process?

A) C₆H₁₂O₆ + 2 ADP + 2 Pi → 2 C₃H₆O₃ + 2 ATP

B) C₆H₁₂O₆ + 2 ATP → 2 C₃H₃O₃⁻ + 2 NADH

C) C₃H₃O₃⁻ + CoA + NAD+ → Acetyl-CoA + CO₂ + NADH

D) Acetyl-CoA + 3 NAD+ + FAD + GDP + Pi → 2 CO₂ + 3 NADH + FADH₂ + GTP

We need a volume of 60.4 ml of 11.7 M HCl to make a 500.0 ml solution with a pH of 3.20.

To calculate the required volume of 11.7 M HCl to make a 500.0 ml solution with a pH of 3.20, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to its pKa and the ratio of the concentrations of the conjugate base and acid.

Using the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻] ÷ [HA])

where [A-] / [HA] is the ratio of the concentration of the conjugate base (Cl⁻) to the concentration of the acid (H⁺).

Rearranging the equation, we can solve for [H⁺]:

[H⁺] = [tex]10^{(pH - pKa)}[/tex]

[H⁺] = [tex]10^{(3.20 - (-1))}[/tex]

= [tex]10^{-3.20} + mol/L[/tex]

Since the concentration of HCl is equal to the concentration of [H⁺] in solution, we can calculate the moles of HCl required to make the solution:

moles of HCl = concentration of HCl × volume of solution

moles of HCl = [tex](10^{-3.20})[/tex] × (0.5 L)

= 7.08 × 10⁻⁴ mol

Finally, we can calculate the required volume of 11.7 M HCl:

volume of HCl = moles of HCl ÷ concentration of HCl

volume of HCl = (7.08 × 10⁻⁴ mol) ÷ (11.7 mol/L)

= 0.0604 L

= 60.4 ml

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Answer:

106.5 J, and it was lost.

Explanation:

To calculate the amount of heat transferred, we can use the following formula:

Q = m * c * ΔT

where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

For graphite, the specific heat capacity is approximately 0.71 J/g°C.

So we have:

Q = 15.0 g * 0.71 J/g°C * (-10°C)

Q = -106.5 J

The negative sign of the answer indicates that the graphite lost heat, since its temperature decreased. Therefore, the heat was transferred from the graphite to its surroundings.

So the amount of heat transferred from the graphite was 106.5 J, and it was lost.

The acceleration due to gravity on the massive planet is 39.48 m/s².

The period (T) of a simple pendulum is given by:

T = 2π√(L/g),

where L is the length of the pendulum and g is the acceleration due to gravity.

In this scenario, we are given that the period of the pendulum (T) is 1 second and the length of the pendulum (L) is 1 meter.

So, substituting these values into the equation:

1 = 2π√(1/g)

Simplifying this equation :

g = (4π²) / (1²)

g = 4π² m/s²

g ≈ 39.48 m/s²

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The pressure of the nitrogen gas at 250.0 K will be 2.78 kPa.

To find the pressure of the nitrogen gas at 250.0 K, we will use the combined gas law formula:
P₁/T₁ = P₂/T₂

Where P₁ is the initial pressure (6.00 kPa), T₁ is the initial temperature (540 K), P₂ is the final pressure (which we want to find), and T₂ is the final temperature (250.0 K).

Since the volume does not change, we can use this simplified formula.

Step 1: Rearrange the formula to solve for P₂:
P₂ = (P₁ × T₂) / T₁

Step 2: Plug in the given values and calculate P₂:
P₂ = (6.00 kPa × 250.0 K) / 540 K

Step 3: Calculate P₂:
P₂ = 1500 / 540 = 2.78 kPa (rounded to two decimal places)

So, the pressure of the nitrogen gas at 250.0 K will be 2.78 kPa.

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The new volume of the soap bubble is approximately 54.29 mL. Since the volume has increased, the bubble will get bigger when the pressure drops to 700.0 mmHg during the thunderstorm.

A 50.0 mL soap bubble is blown at standard pressure. When a thunderstorm passes later in the day, the pressure becomes 700.0 mmHg. To determine if the bubble will get bigger or smaller and to find its new volume, we will use Boyle's Law, which states that P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Step 1: Convert the initial and final pressures to the same unit. The standard pressure is 1 atmosphere (atm), which is equivalent to 760 mmHg. The final pressure is given as 700.0 mmHg.

Step 2: Apply Boyle's Law. Let P1 = 760 mmHg, V1 = 50.0 mL, and P2 = 700.0 mmHg. We will solve for V2, the new volume.

760 mmHg * 50.0 mL = 700.0 mmHg * V2

Step 3: Solve for V2.

V2 = (760 mmHg * 50.0 mL) / 700.0 mmHg
V2 ≈ 54.29 mL

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This attraction is known as hydrogen bonding, which occurs when a hydrogen atom that is covalently bonded to one electronegative atom (such as oxygen) is attracted to another electronegative atom in another molecule. In the case of water molecules, the hydrogen atoms have a partial positive charge and the oxygen atoms have a partial negative charge due to differences in electronegativity. This allows for the formation of hydrogen bonds between adjacent water molecules. The hydrogen bonding gives water its unique properties such as high boiling point and surface tension.

Tectonic events can impact the flow of radiant energy in various ways. One of the primary ways is through the formation of mountains and the alteration of landforms.

When tectonic plates collide and push against each other, they can form mountains, which can affect the flow of radiant energy. Mountains can block or redirect the flow of wind, which in turn can affect the amount of solar radiation that reaches the earth's surface.

They can also create changes in atmospheric pressure and temperature that impact the movement of air masses, which can affect the flow of radiant energy.

Tectonic events can also impact the flow of radiant energy by altering the composition of the atmosphere.

For example, volcanic eruptions can release large amounts of sulfur dioxide and other particles into the atmosphere, which can reflect and scatter incoming solar radiation, leading to cooling of the earth's surface.

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Saccharin, an artificial sweetener that is 3000 times sweeter than sucrose, is composed of a) C₁₄H₁₀O₆N₂S₂.

45.90% carbon, 2.73% hydrogen, 26.23% oxygen, 7.65% nitrogen, and 17.49% sulfur. is the molecular formula of saccharin.

To determine the molecular formula of saccharin, we first need to calculate the empirical formula using the given percentages of each element.

Assuming we have 100 grams of saccharin, we have:

Carbon: 45.90 g / 12.01 g/mol = 3.82 mol

Hydrogen: 2.73 g / 1.01 g/mol = 2.70 mol

Oxygen: 26.23 g / 16.00 g/mol = 1.64 mol

Nitrogen: 7.65 g / 14.01 g/mol = 0.55 mol

Sulfur: 17.49 g / 32.07 g/mol = 0.55 mol

We can divide each value by the smallest one, which is 0.55 mol, to get the following ratios:

Carbon: 3.82 / 0.55 = 6.95

Hydrogen: 2.70 / 0.55 = 4.91

Oxygen: 1.64 / 0.55 = 2.98

Nitrogen: 0.55 / 0.55 = 1

Sulfur: 0.55 / 0.55 = 1

The resulting ratios are close to whole numbers, so we can assume the empirical formula to be C₇H₅NO₃S. To find the molecular formula, we need to determine the actual molecular mass of saccharin.

The empirical formula mass of C₇H₅NO₃S is approximately 183 g/mol. The molecular mass of saccharin is known to be around 452 g/mol, so we can calculate the ratio of the molecular mass to the empirical formula mass:

452 g/mol / 183 g/mol = 2.47

This means that the molecular formula is 2.47 times the empirical formula, or:

C₇H₅NO₃S * 2.47 = C₁₇H₁₃N₂O₅S

Therefore, the molecular formula of saccharin is (a) C₁₄H₁₀O₆N₂S₂. The other options (b) CSH,ONS, (c) C&H₉O₂NS, and (d) C;H₅O₃NS are not correct.

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To determine the formula of the hydrated salt, we need to first find the empirical formula by determining the smallest whole number ratio of the elements present in the compound.

Then, we can use the molar mass of the empirical formula and the percentage composition of the water to find the molecular formula.

Step 1: Find the empirical formula

Assuming 100 g of the compound, we can calculate the masses of each element present:

- Iron: 20.14 g

- Oxygen: 23.02 g

- Sulphur: 11.51 g

- Water: 45.32 g

Next, we need to convert these masses to moles:

- Iron: 20.14 g / 55.85 g/mol = 0.360 mol

- Oxygen: 23.02 g / 16.00 g/mol = 1.439 mol

- Sulphur: 11.51 g / 32.06 g/mol = 0.359 mol

- Water: 45.32 g / 18.02 g/mol = 2.515 mol

We can then divide each mole value by the smallest mole value to get the mole ratio:

- Iron: 0.360 mol / 0.359 mol ≈ 1

- Oxygen: 1.439 mol / 0.359 mol ≈ 4

- Sulphur: 0.359 mol / 0.359 mol = 1

- Water: 2.515 mol / 0.359 mol ≈ 7

The mole ratio is approximately 1:4:1:7, which gives us the empirical formula:

FeSO4·7H2O

Step 2: Find the molecular formula

The empirical formula mass of FeSO4·7H2O is:

(55.85 + 32.06 + 4(16.00)) + 7(18.02) = 278.00 g/mol

We know from the problem that the molecular mass of the salt is 278 g/mol, so the empirical formula is also the molecular formula. Therefore, the formula of the hydrated salt is FeSO4·7H2O.

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Mg₃N₂ will be completely consumed, and there will be some H₂O left over after the reaction is complete.

To determine the limiting reactant, we need to compare the number of moles of each reactant present to the stoichiometric ratio in the balanced equation.

From the balanced equation, we see that for every 1 mole of Mg₃N₂, 6 moles of H₂O are required. Therefore, the stoichiometric ratio of Mg₃N₂ to H₂O is 1:6.

To find out which reactant is limiting, we can calculate the amount of products that each reactant could produce.

For Mg₃N₂:
0.600 mol Mg₃N₂ x (3 mol Mg(OH)₂ / 1 mol Mg₃N₂) = 1.80 mol Mg(OH)₂

For H₂O:
4.00 mol H₂O x (3 mol Mg(OH)₂ / 6 mol H₂O) = 2.00 mol Mg(OH)₂

Since Mg₃N₂ can only produce 1.80 mol Mg(OH)₂, which is less than the amount that H₂O can produce (2.00 mol), Mg₃N₂ is the limiting reactant.

Therefore, Mg₃N₂ will be completely consumed, and there will be some H₂O left over after the reaction is complete.

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I₂ is a solid, Br₂ is a liquid, while Cl₂ and F₂ are gases because of their increasing molecular size and decreasing strength of their intermolecular forces.

The main factor influencing the physical states of halogens is the strength of the intermolecular forces (Van der Waals forces) between their molecules.

As you move down Group 17 in the periodic table (from F₂ to I₂), the size and mass of the halogen molecules increase. Larger molecules have a greater number of electrons, leading to stronger dispersion forces (a type of Van der Waals forces) between molecules.

For I₂, these forces are strong enough to hold the molecules together in a solid form. For Br₂, the forces are slightly weaker but still strong enough to form a liquid. However, in Cl₂ and F₂, the forces are weaker, allowing the molecules to be in a gaseous state at room temperature.

In summary, the physical states of the halogens depend on the strength of their intermolecular forces, which is influenced by the size and mass of the molecules.

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a. HNO3 + KOH --> KNO3 + H2O

b. HCl + LiOH --> LiCl + H2O

c. H2S + 2NaOH --> Na2S + 2H2O

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In the following acid-base reaction, hpo₄²⁻ is the base.

This can be seen as it accepts a proton (H⁺) from H₂O to form the conjugate acid, H₂PO₄⁻. The other reactant, H₂O, donates the proton, making it the acid in the reaction. It is important to note that in an acid-base reaction, the species that donates a proton is the acid and the species that accepts the proton is the base.

The strength of the acid and base can also be determined by the equilibrium constant of the reaction. The larger the equilibrium constant, the stronger the acid or base. In this particular reaction, hpo₄²⁻ is a weak base, as it only partially accepts the proton from H₂O.

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a) A pH of 5 for river water near a factory does suggest a potential pollution problem. The normal pH range for most natural waters is around 6.5-8.5. pH values below 6.5 can indicate acidification, which can be caused by pollutants such as sulfur dioxide and nitrogen oxides from industrial activities, or from natural sources such as acid rain.

A pH of 5 is more acidic than most natural waters and could indicate the presence of acidic pollutants in the water.

Therefore, in terms of b) Other evidence that would be useful to consider before reaching a conclusion about whether the pH of 5 represents a pollution problem includes:

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The mass of copper sulfate in the given solution is 24 grams.

Copper sulfate, also known as cupric sulfate or copper (II) sulfate, is a chemical compound that consists of copper ions and sulfate ions. It has the molecular formula CuSO4 and is commonly used in agriculture, mining, and chemical industries.

In the given scenario, we have a solution of copper sulfate with a volume of 2 dm3 and a concentration of 12 g/dm3. This means that for every 1 dm3 of the solution, there are 12 grams of copper sulfate present. To find the mass of copper sulfate in the entire 2 dm3 solution, we can use the following formula:

Mass = Concentration x Volume

Substituting the given values, we get:

Mass = 12 g/dm3 x 2 dm3
Mass = 24 g

Therefore, the mass of copper sulfate in the given solution is 24 grams.

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The equilibrium constant, K, for the reaction H2(g) + I2(g) ⇄ 2HI(g) can be calculated using the expression K= [HI]2/([H2][I2]). Since the concentrations of H2, I2, and HI are given in the question, we can calculate the equilibrium constant, K, for the reaction.

K = [HI]2/([H2][I2]) = (1.77 × 10-2)2/((1.83 × 10-3)(3.13 × 10-3)) = 4.43 × 104. Therefore, the equilibrium constant, K, for the reaction H2(g) + I2(g) ⇄ 2HI(g) at 425°C is 4.43 × 104.

Using the specified concentrations of H2, I2, and HI, it appears that you have correctly calculated the equilibrium constant, K, for the reaction H2(g) + I2(g) 2HI(g) at 425°C. The ratio of the concentrations of the reactants and products at equilibrium, K, is represented by each concentration being raised to the power of its stoichiometric coefficient.

The concentration of the product, HI, is preferred above the concentrations of the reactants, H2 and I2, at equilibrium, as shown by the value of K = 4.43 104 in this instance. This suggests that at equilibrium, the forward reaction—the creation of HI—is preferred.

It is significant to remember that the equilibrium constant, K, is temperature-dependent, and that temperature changes affect K's value.

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If we add 45.0 g of sodium chloride to 500.0 g of water,  the melting point is -  5.7 °C and the boiling point of the solution is 101.5 °C.

The mass of the NaCl = 45 g

The mass of the water = 500 g

The moles of the NaCl = mass / molar mass

                                      = 45 / 58.44

                                      = 0.770 mol

The molality is expressed as :

b = moles of solute / mass of solvent in kg

b = 0.770 / 0.5

b = 1.54 m

The boiling-point elevation  :

ΔTb = 2 × 0.512 × 1.54

       = 1.5 ° C

The boiling point, Tb = 100°C + 1.5 °C

                                   = 101.5 °C

The expression is as :

ΔTf = 2 × 1.86 × 1.54

     = 5.7 °C

The melting point = 0 - 5.7

                              = - 5.7 °C

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We need 0.4 L of the 0.5 M MgSO₄ stock solution to make 2.0 L of 0.20 M MgSO₄.

To calculate the volume of the 0.5 M MgSO₄ stock solution needed to make 2.0 L of 0.20 M MgSO₄, we will use the formula m₁v₁ = m₂v₂.

1. Identify the given values:
m₁ = 0.5 M (concentration of the stock solution)
m₂ = 0.20 M (concentration of the desired solution)
v₂= 2.0 L (volume of the desired solution)

2. Plug the given values into the formula:
(0.5 M)(v₁) = (0.20 M)(2.0 L)

3. Solve for v1 (volume of the stock solution needed):
v₁= (0.20 M)(2.0 L) / (0.5 M)
v₁= 0.4 L

So, you need 0.4 L of the 0.5 M MgSO₄ stock solution to make 2.0 L of 0.20 M MgSO₄.

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