CHEMISTRY chemistry Table balance A+B→C Table2

1. To solve for the grams of H2 needed, we need to use the given ΔH value to calculate the amount of moles of N2 that reacted. From the balanced chemical equation, we know that for every 3 moles of H2 that reacts, 1 mole of N2 reacts. Therefore, we can use the mole ratio to convert the moles of N2 to moles of H2 and then use the molar mass of H2 to convert to grams.

First, we need to calculate the moles of N2 that reacted to produce 150.9kJ of heat:
ΔH = -92.40 kJ/mol N2
150.9 kJ = (1 mol N2 / -92.40 kJ) x (-150.9 kJ)
mol N2 = 1.63 mol

Using the mole ratio from the balanced chemical equation:

1 mol N2 : 3 mol H2

We can calculate the moles of H2 needed:3 mol H2 = 1 mol N2

3 mol H2 = 1.63 mol N2
mol H2 = 0.543 mol

Finally, we can convert moles of H2 to grams:

mol H2 = 0.543 mol
molar mass of H2 = 2.02 g/mol
grams of H2 = (0.543 mol) x (2.02 g/mol)
grams of H2 = 1.10 g

Therefore, 1.10 grams of H2 are needed to involve 150.9kJ of heat.

2. To solve for the moles of NH3 produced, we can use the same mole ratio from the balanced chemical equation:

1 mol N2 : 2 mol NH3

From the moles of N2 that reacted calculated in part 1, we can calculate the moles of NH3 produced:
1 mol N2 = 2 mol NH3
1 mol N2 = 1.63 mol N2
mol NH3 = (2 mol NH3 / 1 mol N2) x (1.63 mol N2)
mol NH3 = 3.26 mol

Therefore, 3.26 moles of NH3 were produced in the process.

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

The following data were obtained from the quesion:

The equilibrium constant for the reaction can be obtain as illustrated below:

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

Equilibrium constant = (0.007 × 0.0118) / 0.0168

Equilibrium constant = 0.0049

Thus, from the above calculation, the equilibrium constant, K for the reaction is 0.0049

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As the hour of her new job approached, Emma could feel her excitement peak.

"Peak" is a synonym to "intensify" in this context because it means to reach the highest point or level of something. In the given passage, Emma's excitement is growing stronger and stronger as the time for her volunteer job approaches.

When her excitement "peaks," it means that it has reached the highest point of intensity, just like when something is intensified, it becomes stronger or more intense.

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Answer:

Leaves are green due to the presence of an organelle chloroplast (in abundance) which contains the pigment chlorophyll

Explanation:

Now saying chlorophyll pigment  is a green pigment might be slightly incorrect. The two famous types (Chlorophyll a, Chlorophyll b) only absorb red and blue light from the atmosphere and reflect green light hence giving the pigment a green appearance and lastly giving the leaves a green color too

Answer:

Chlorophyll

Explanation:

Plants are often seen as green to the human eye due to the presence of chlorophyll, which is the primary pigment used in photosynthesis. Chlorophyll absorbs light in the red and blue-violet parts of the spectrum, but reflects or transmits green light, resulting in the characteristic green color of leaves.

The molar enthalpy for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) is 266.4 kJ/mol.

The molar enthalpy is determined from Hess's law as follows:

Equation 1 x2:

2 N₂ (g) + 2 O₂ (g) → 4 NO (g) ∆H° = 361.2 kJ/mol

Equation 3 x3, :

6 H₂ (g) + 3 O₂ (g) → 6 H₂O (g) ∆H° = -1451.1 kJ/mol

Equation 2 x -4:

-8 N₂ (g) - 12 H₂ (g) → -8 NH₃ (g) ∆H° = 367.2 kJ/mol

Adding the equations together:

-6 N₂ (g) - 6 H₂ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) - 8 NH₃ (g) ∆H° = 266.3 kJ/mol

Multiplying the equation above by -1/2:

3 N₂ (g) + 3 H₂ (g) - 5/2 O₂ (g) → -2 NO (g) - 3 H₂O (g) + 4 NH₃ (g) ∆H° = -133.2 kJ/mol

Multiplying the above equation by -2:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = 266.4 kJ/mol

This is the molar enthalpy of the given reaction

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Boiling point of solution will be 100 C. The incorrect statement is C)

An aqueous solution is one in which water serves as the solvent. One or more substances are dissolved in water to create such a solution, and the water molecules surround and separate the individual solute particles to create a homogeneous mixture.

Therefore, A solvent's boiling point and freezing point change when a solute is dissolved in it, respectively. Boiling point elevation and freezing point depression are two terms used to describe this occurrence. The concentration of the solute determines how much of an impact it has.

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Answer:

2.000 moles.

Explanation:

To solve this question, we need to use the formula:

n = V / VM

where n is the number of moles, V is the volume of water in milliliters, and VM is the molar volume of water in milliliters per mole. The molar volume of water at standard temperature and pressure (STP) is 18.02 mL/mol. Plugging in the given value, we get:

n = 36.030 mL / 18.02 mL/mol n = 2.000 moles

The answer is 2.000 moles.

Answer:

649.090

Explanation:

Answer:

A solvent front is the point on a chromatography paper or plate where the solvent has reached the end of the stationary phase and has migrated as far as it can go. It is the farthest point reached by the solvent in the chromatography process.

the oxidation number of copper in copper oxide is 2...

Answer: +2

Explanation: Copper has a +2 oxidation number in CuO.

This is due to the fact that oxygen has an oxidation number of 2, and the entire chemical has a neutral charge. Consequently, the following equation can be used to determine copper's oxidation number:

(+2) + (-2) = 0

In order to counteract the -2 oxidation number of oxygen in CuO, copper must have an oxidation number of +2.

The decay of Xenon-146 follows an exponential decay model, where the amount of Xenon-146 remaining after time t is given by:

N(t) = N0 * (1/2)^(t/T)

where N0 is the initial amount of Xenon-146, T is the half-life of Xenon-146, and t is the time that has passed.

We are given that N0 = 416 g, N(t) = 13 g, and T = 37 hours. We can solve for t as follows:

N(t) = N0 * (1/2)^(t/T)

13 = 416 * (1/2)^(t/37)

Taking the natural logarithm of both sides:

ln(13) = ln(416) + (t/37) * ln(1/2)

Solving for t:

t = 37 * [ln(13/416) / ln(1/2)]

t ≈ 111.2 hours

Therefore, the Xenon-146 has been decaying for approximately 111.2 hours.

1. The long-haired cat in the P generation is a purebred. This is because it has two copies of the recessive allele (hh) responsible for long hair.

2. The short-haired cat in the P generation is a hybrid. We know this because the offspring include both short-haired (Hh) and long-haired (hh) cats, indicating that the short-haired parent must have one dominant (H) and one recessive (h) allele (Hh).

3. If the short-haired cat in the P generation were purebred (HH), all offspring would have short hair, as they would inherit one dominant allele (H) from the short-haired parent and one recessive allele (h) from the long-haired parent, resulting in Hh offspring.

4. The black horse is a hybrid. Since the cross between a black horse (B...) and a brown horse (bb) produced a brown foal (bb), the black horse must carry one dominant allele (B) and one recessive allele (b) - making it a hybrid (Bb).

5. To determine whether a guinea pig with a smooth coat (S...) is a hybrid or a purebred, perform a test cross by mating it with a guinea pig with a rough coat (ss). If all offspring have smooth coats (Ss), the smooth-coated guinea pig is likely purebred (SS). If any offspring have a rough coat (ss), the smooth-coated guinea pig is a hybrid (Ss).

A dominant allele is a variant of a gene that expresses its trait even when only one copy is present in an individual's genotype. In other words, it masks the effect of another variant (allele) of the same gene when they are together.

A recessive allele is a variant of a gene that only expresses its trait when two copies are present in an individual's genotype. The trait associated with the recessive allele is "masked" by the presence of a dominant allele, and it will only be expressed if both copies of the gene are recessive.

The above answer is based on the question below;

In a test cross, the organism with the trait controlled by a dominant allele is crossed with an organism with a trait controlled by a recessive allele. If all offspring have the trait controlled by the dominant allele, then the parent is probably a purebred. If any offspring has the recessive strait, then the dominant parent is a hybrid.

1. Is the long-haired cat in the P generation a hybrid or a purebred? Explain your answer.

2. Is the short-haired cat in the P generation a hybrid or a purebred? Explain your answer.

3. If the short-haired cat in the P generation were purebred, what would you expect the offspring to look like?

4. In horses, the allele for a black coat (B) is dominant over the allele for a brown coat (b). A cross between a black horse and a brown horse produces a brown foal. Is the black horse a hybrid or a purebred? Explain.

5. In guinea pigs, the allele for a smooth coat (S) is dominant over the allele for a rough coat (s). Explain how you could find out whether a

guinea pig with a smooth coat is a hybrid or a purebred.

H= Short hair

h = Long hair

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A silver bar with the mass of the 300 grams is heated from the 30 °C to 55 °C. The amount heat does the silver bar absorb in the joules is 1762.5 J.

The mass of the silver bar = 300 g

The initial temperature = 30 °C

The final temperature = 55 °C

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the silver bar = 300 g

The c is the specific heat capacity = 0.235 J/g °C

The ΔT is the change in the temperature = final temperature - initial temperature

The ΔT is the change in the temperature = 55 °C - 30 °C

The ΔT is the change in the temperature = 25 °C

The heat energy, Q = 300 × 0.235 × 25

The heat energy, Q = 1762.5 J

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The amount of heat required to vaporize 181.20 g of ethanol would be 170.1 kJ.

Using the formula:

Q = n ∆Hvap

where:

Moles of 181.20 g of ethanol = 181.20 g / 46.07 g/mol = 3.933 mol

Substituting the values:

Q = 3.933 mol x 43.3 kJ/mol = 170.1 kJ

In other words, the amount of heat required to vaporize 181.20 g of ethanol is 170.1 kJ.

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a) The amount of N₂O₅ is lowered to 2.5 g during the course of around 4.41 × 10⁴  seconds or 12.25 hours.

b) 9.71 L of O₂ are generated at 745 mmHg and 45 °C.

a) To solve for the time required for the quantity of N₂O₅ to be reduced to 2.5 g, use the first-order integrated rate law:

ln[N₂O₅]t/[N₂O₅]0 = -kt

where [N₂O₅]t = concentration of N₂O₅ at time t, [N₂O₅]0 = initial concentration of N₂O₅, k = rate constant, and t = time.

Find the initial concentration of N₂O₅:

n(N₂O₅) = m/M = 80.0 g / 108.01 g/mol = 0.7413 mol

[N₂O₅]0 = n/V = 0.7413 mol / 0.153 L = 4.846 M

where M = molar mass of N₂O₅ and V = volume of the solution.

Substituting the given values into the equation:

ln([N₂O₅]t / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

When the quantity of N₂O₅ is reduced to 2.5 g, the concentration is:

n(N₂O₅) = m/M = 2.5 g / 108.01 g/mol = 0.02314 mol

[N₂O₅]t = n/V = 0.02314 mol / 0.153 L = 0.151 M

Substituting this concentration into the equation and solving for t:

ln(0.151 M / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

t = 4.41 × 10⁴ s

Therefore, it takes approximately 4.41 × 10⁴ seconds or 12.25 hours for the quantity of N₂O₅ to be reduced to 2.5 g.

b) The balanced equation for the reaction shows that 1 mole of N₂O₅ produces 1/2 mole of O₂:

N₂O₅ → N₂O₄ + 1/2 O2

Therefore, the number of moles of O₂ produced can be calculated using the stoichiometry:

n(O₂) = 1/2 × n(N₂O₅) = 1/2 × 0.7413 mol = 0.3707 mol

The ideal gas law can be used to calculate the volume of O₂ produced at 745 mmHg and 45°C:

PV = nRT

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

Convert the pressure to atm and the temperature to Kelvin:

P = 745 mmHg / 760 mmHg/atm = 0.980 atm

T = 45°C + 273.15 = 318.15 K

Substituting the values and solving for V:

V = nRT/P = (0.3707 mol) × (0.08206 L·atm/mol·K) × (318.15 K) / (0.980 atm) = 9.71 L

Therefore, the volume of O₂ produced at 745 mmHg and 45°C is 9.71 L.

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Answer: Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as

Explanation:

The limiting reagent for the reaction between 0.150 g of salicylic acid and 0.350 mL of acetic anhydride is salicylic acid, C₇H₆O₃

First, we shall determine the mass of the acetic anhydride. Details below:

Mass = density × volume

Mass of acetic anhydride, C₄H₆O₃ = 1.082 × 0.350

Mass of acetic anhydride, C₄H₆O₃ = 0.3787 g

Finally, we shall determine the limiting reagent. Details below:

C₇H₆O₃ + C₄H₆O₃ -> C₉H₈O₄ + CH₃COOH

From the balanced equation above,

138.121 g of C₇H₆O₃ reacted with 102.09 g of C₄H₆O₃

Therefore,

0.150 g of C₇H₆O₃ will react with = (0.150 × 102.09) / 138.121 = 0.11089 g of C₄H₆O₃

We can see from the above that only 0.11089 g of acetic anhydride, C₄H₆O₃ out of 0.3787 g is needed to react with 0.150 g of salicylic acid, C₇H₆O₃

Thus, the limiting reagent is salicylic acid, C₇H₆O₃

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The height of fall of the ice is determined as 313.25 m.

The total heat energy of the ice during the fall is calculated as follows;

Q = ml + mcΔT

where;

Q = 334000 x 0.4  +  0.4 x 4200 x (10)

Q = 150,400 J

The energy converted into potential energy is calculated as;

3%K.E = 150,400 J

0.03K.E = 150,400 J

K.E = 5,013,333.33 J

¹/₂mv² = 5,013,333.33 J

v = √(2 x 5,013,333.33)/(0.4)

v = 5,006.67 m/s

The height of fall is calculated as;

h = √2gh

h = √(2 x 5,006.67 x 9.8)

h = 313.25 m

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The volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.

A mole is a unit of measurement used in chemistry to represent the quantity of a chemical.

We can use the following formula to get the volume of a 1.0 M solution containing 4.0 moles of solute:

moles of solute = molarity x volume of solution

To determine the volume of the solution, we can rearrange this formula as follows:

Volume of solution = molarity / moles of solute.

By entering the specified values, we obtain:

4.0 moles / 1.0 M is the solution's volume.

Solution volume = 4.0 L

Therefore, the volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.

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The S-P difference (sec) is used to calculate the distance (km) between an earthquake epicenter and a seismic station, while the magnitude (M) is a measure of the energy released during the earthquake.

These parameters are important for understanding the severity and impact of an earthquake, as well as for predicting future seismic activity.

The S-P difference (sec) refers to the time difference between the arrival of the primary (P) waves and the secondary (S) waves at a seismic station. This time difference is used to calculate the distance (km) between the earthquake epicenter and the seismic station, using the equation: distance (km) = S-P difference (sec) x 8 km/sec. This calculation assumes that the waves travel at a constant speed through the Earth's interior.
The magnitude (M) of an earthquake is a measure of the energy released during the earthquake, and is usually determined using a seismometer. The magnitude scale is logarithmic, meaning that each increase of one unit represents a tenfold increase in seismic energy. For example, an earthquake with a magnitude of 5.0 is ten times more powerful than one with a magnitude of 4.0, and 100 times more powerful than one with a magnitude of 3.0.

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The volume of oxygen gas, O₂ collected at 1.05 atm pressure and 44.0 °C when 42.5 g of KClO₃ decomposed is 13.01 L

We shall begin by obtaining the mole in 42.5 g of KClO₃. Details below:

Mole = mass / molar mass

Mole of CaC₂ = 42.5 / 122.5

Mole of CaC₂ = 0.35 mole

Next, we shall determine the mole of oxygen gas, O₂. produced. Details below:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produced 3 mole of O₂

Therefore,

0.35 mole of KClO₃ will decompose to produce = (0.35 × 3) / 2 = 0.525 mole O₂

Finally, we shall determine the volume of oxygen gas, O₂ collected. Details below:

PV = nRT

1.05 × V = 0.525 × 0.0821 × 317

Divide both sides by 1.05

V = (0.525 × 0.0821 × 317) / 1.05

Volume of oxygen gas = 13.01 L

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Hydrazine, reacts with the oxygen to form the nitrogen gas and the water. The percent yield of the reaction is 3.18 %.

The balanced reaction is :

N₂H₄  + O₂  --->  N₂ + 2H₂O

The mass of the N₂H₄  = 3.55 g

The moles of N₂H₄ = mass / molar mass

The moles of N₂H₄ = 3.55 / 32

The moles of N₂H₄ = 0.110 mol

The theoretical yield = 0.110 mol × 28 g/mol

The theoretical yield = 3.08 g

The gas equation is :

P V = n R T

n = P V / R T

n = (1 × 0.850 ) / ( 0.0823 ×295 )

n = 0.0035 mol

The actual yield = 0.0035 × 28

The actual yield = 0.098 g

The percent yield = ( 0.098 / 3.08 ) × 100 %

The percent yield = 3.18 %.

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Under the specified conditions, the cell potential of the given galvanic cell is 3.11 V.

The standard reduction potentials for the half-reactions involved in the cell reaction are:

Ti²⁺(aq) + 2e- ⇆ Ti(s) E° = -1.63 V

Au³⁺(aq) + 3e- ⇆ Au(s) E° = +1.50 V

The cell potential (Ecell) is given by:

Ecell = E°cathode - E°anode

where E°cathode = standard reduction potential of the cathode (reduction half-reaction) and E°anode = standard reduction potential of the anode (oxidation half-reaction).

In this case, Ti²⁺ is oxidized (anode) and Au³⁺ is reduced (cathode). Therefore:

E°anode = -1.63 V

E°cathode = +1.50 V

So, Ecell = +1.50 V - (-1.63 V) = +3.13 V

The Nernst equation can be used to calculate the cell potential (Ecell) under non-standard conditions:

Ecell = E°cell - (RT/nF) ln(Q)

where R = gas constant (8.314 J/(molK)), T = temperature in Kelvin (25 °C = 298 K), n = number of electrons transferred in the balanced equation (3 in this case), F = Faraday constant (96,485 C/mol), and Q = reaction quotient.

For the given concentrations:

[Ti²⁺] = 0.00140 M

[Au³⁺] = 0.887 M

The reaction quotient Q can be written as:

Q = ([Ti²⁺]³/[Au³⁺]²)

Substituting the values into the Nernst equation:

Ecell = E°cell - (RT/nF) ln([Ti²⁺]³/[Au³⁺]²)

Ecell = 3.13 V - (8.314 J/(molK) × 298 K / (3 × 96,485 C/mol)) ln(0.00140³/0.887²)

Ecell = 3.13 V - 0.0217 V

Ecell = 3.11 V

Therefore, the cell potential for the given galvanic cell under the given conditions is 3.11 V.

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The identity of element X in the compound X2Fe(CN)6 · 12H2O is sodium (Na).

To find the identity of element X in the compound X2Fe(CN)6 · 12H2O, we can start by determining the molar mass of the compound.

The molar mass of X2Fe(CN)6 is:

2 × molar mass of X + molar mass of Fe + 6 × molar mass of C + 6 × molar mass of N

= 2 × atomic mass of X + atomic mass of Fe + 6 × 12.01 g/mol + 6 × 14.01 g/mol

= 2 × atomic mass of X + 55.85 g/mol + 432.72 g/mol + 84.06 g/mol

= 2 × atomic mass of X + 572.63 g/mol

The molar mass of 12H2O is:

12 × (atomic mass of H + atomic mass of O) = 12 × (1.01 g/mol + 16.00 g/mol) = 216.24 g/mol

The total molar mass of the compound is:

2 × atomic mass of X + 572.63 g/mol + 216.24 g/mol = 2 × atomic mass of X + 788.87 g/mol

Now we can use the given information that the compound is 45.34% water by mass. This means that the mass of water in the compound is 45.34% of the total mass of the compound, and the mass of the rest of the compound (X2Fe(CN)6) is 100% - 45.34% = 54.66% of the total mass of the compound.

Let's assume we have 100 g of the compound. Then the mass of water in the compound is:

45.34 g water = 0.4534 × 100 g compound

The mass of the rest of the compound (X2Fe(CN)6) is:

54.66 g rest of the compound = 0.5466 × 100 g compound

We can now use the mass of the rest of the compound (X2Fe(CN)6) to find the number of moles of the compound:

moles of X2Fe(CN)6 = (54.66 g) / (2 × atomic mass of X + 572.63 g/mol)

We can also use the mass of water to find the number of moles of water:

moles of H2O = (45.34 g) / 18.02 g/mol

Since the compound has 12 moles of water per mole of X2Fe(CN)6, we have:

moles of X2Fe(CN)6 = 1/12 × moles of H2O

We can now set these two expressions for moles of the compound equal to each other and solve for the atomic mass of X:

(54.66 g) / (2 × atomic mass of X + 572.63 g/mol) = 1/12 × (45.34 g) / 18.02 g/mol

Simplifying this equation and solving for the atomic mass of X gives:

atomic mass of X = 22.99 g/mol

The atomic mass of X is very close to the atomic mass of sodium (22.99 g/mol), so it is likely that X is sodium. Therefore, the identity of element X in the compound X2Fe(CN)6 · 12H2O is sodium (Na).

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5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

Molarity (M) is defined as the moles of solute per liter of the solution.

Balanced chemical equation is : 2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl₃ and 3 moles of H₂.

As moles of HCl = Molarity × Volume

moles of HCl = 1.65 mol/L × 0.352 L

moles of HCl = 0.58128 mol

and moles of Al = (2/6) × moles of HCl

moles of Al = (1/3) × 0.58128 mol

moles of Al = 0.19376 mol

mass of Al = moles of Al × molar mass of Al

mass of Al = 0.19376 mol × 26.98 g/mol

mass of Al = 5.221 g

So, 5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

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If a skimmer and a gull eat the same fish and the skimmer is more successful at catching the fish over time, it is likely that the skimmer population would increase, while the gull population may decrease.

The skimmer's success in catching the fish would give it an advantage in obtaining the necessary nutrients for survival and reproduction. As a result, the skimmer population would likely grow over time as more individuals are able to survive and reproduce due to the abundance of food.

On the other hand, the gull population may decrease due to the competition with the skimmer for the same food source. If the skimmer population grows significantly, it may lead to a reduction in the availability of fish for the gulls to feed on. Over time, this could result in a decline in the gull population due to reduced food availability.

However, it is important to note that the impact on the bird populations may depend on various factors such as the size of the populations, availability of other food sources, and environmental factors. Therefore, the outcome of this scenario cannot be predicted with certainty and would require further analysis and investigation.

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The given reaction equation tells us that for every 1 mol of C₅H₁₂, 5 moles of CO₂ will be produced. Since 8.25 mol of C₅H₁₂ is given, 8.25 mol C₅H₁₂ x 5 moles CO₂/1 mol C₅H₁₂ = 41.25 moles CO₂ will be produced.

Reaction is the process of responding to an event or stimulus in a particular way. It can occur at the physical, cognitive, or emotional level. Physically, a reaction could be as simple as a reflex or as complex as a multi-step process. Cognitively, it could involve forming a judgment or understanding. Emotionally, it could involve feelings of fear, shock, anger, or joy. In the context of science, reactions are often chemical or physical processes that involve the conversion of one set of substances into another.

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Answer:

56.11 g/mol

Explanation:

To determine the molar mass of potassium hydroxide, we need to find the atomic mass of each element in the compound and add them up.

The atomic mass of potassium (K) is 39.10 g/mol, the atomic mass of oxygen (O) is 16.00 g/mol, and the atomic mass of hydrogen (H) is 1.01 g/mol.

So, the molar mass of potassium hydroxide (KOH) is:

Molar mass of K = 39.10 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of KOH = Molar mass of K + Molar mass of O + Molar mass of H

= 39.10 g/mol + 16.00 g/mol + 1.01 g/mol

= 56.11 g/mol

Therefore, the molar mass of potassium hydroxide (KOH) is 56.11 g/mol.

Answer:

Explanation:

We can use the formula:

q = mcΔT

where q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Given:
specific heat of iron (c) = 0.449 J/g°C
initial temperature (T1) = 24.0°C
final temperature (T2) = 82.1°C
heat absorbed (q) = 948.0 J

Substituting the given values into the formula, we get:

q = mcΔT
948.0 J = m(0.449 J/g°C)(82.1°C - 24.0°C)
948.0 J = m(0.449 J/g°C)(58.1°C)
m = 948.0 J ÷ (0.449 J/g°C × 58.1°C)
m = 33.1 g

Therefore, the mass of the iron piece is 33.1 g (to three significant figures)