# In a calorimetry experiment, it was determined that a 92.0 gram piece of copper metal released 1860 J of heat to the surrounding water in the calorimeter (qcopper = −1860 J). If the final temperature of the copper metal-water mixture was 25.00°C, what was the initial temperature of the copper metal? The specific heat of copper is 0.377 J/(g°C). Group of answer choices Tinitial = 28.6°C Tinitial = −28.6°C Tinitial = 78.6°C Tinitial = 92.6°C Tinitial = 53.6°C

Tinitial = 78.6°C

Explanation:

In a calorimetry experiment, the flow heat is measured for a system that has a state change. In this case, there isn't happening a physical change, so the heat is called sensitive heat, and it's calculated by:

q = mxCpxΔT

Where q is the heat, m is the mass, Cp is the specific heat and ΔT is the difference of final and initial temperature (Tfinal - Tinitial).

Copper is losing heat, so q is negative, then:

-1860 = 92x0.377x(25 - Ti)

34.684(25 - Ti) = -1860

25 - Ti = -53.63

-Ti = -78.63

Ti = 78.6ºC

## Related Questions

Which unit would be most appropriate for measuring the volume of water in a swimming pool? Liters or Kiloliters

The most appropriate for measuring the volume of water in a swimming pool is kilolitres.

Consider a pool that is 25 m long by 10 m wide with an average depth of 1.5 m.

V = lwh = 25 m × 10 m × 1.5 m = 375 m³

1 L = 1 dm³, so

V = 375 m³ × (10 dm/1 m)³ = 375 000 dm³ = 375 000 L = 375 kL

Thus, the cubic metre or kilolitre is more appropriate, because it gives more manageable numbers (i.e., between 0.1 and 1000)

Complete combustion of 7.80 g of a hydrocarbon produced 25.1 g of CO2 and 8.55 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary

The empirical formula is C3H5

Explanation:

Step 1: Data given

Mass of the compound = 7.80 grams

Mass of CO2 = 25.1 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 8.55 grams

Molar mass of H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 25.1 grams / 44.01 g/mol

Moles CO2 = 0.570 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.570 moles CO2 we have 0.570 moles C

Step 4: Calculate mass C

Mass C = 0.570 moles * 12.01 g/mol

Mass C = 6.846 grams

Step 5: Calculate moles H2O

Moles H2O = 8.55 grams / 18.02 g/mol

Moles H2O = 0.474 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.474 moles H2O we have 2*0.474 = 0.948 moles H

Step 7: Calculate mass H

Mass H = 0.948 moles * 1.01 g/mol

Mass H = 0.957 grams

Step 8: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.570 moles / 0.570 = 1

H: 0.948 moles / 0.570 = 1.66

This means for 1 mol C we have 1.66 moles H   OR for 3 moles C we have 5 moles H

The empirical formula is C3H5

To find the empirical formula of the hydrocarbon, divide the moles of CO2 and H2O by their molar masses. Use the smallest mole ratio to determine the empirical formula.

### Explanation:

To find the empirical formula of the hydrocarbon, we need to determine the mole ratios between carbon and hydrogen in the compound. First, calculate the moles of CO2 produced by dividing the mass of CO2 by its molar mass. Next, calculate the moles of H2O produced by dividing the mass of H2O by its molar mass. Finally, divide the moles of each element by the smallest number of moles to obtain the mole ratio between carbon and hydrogen. The empirical formula is CnHm, where n and m represent the mole ratios of carbon and hydrogen, respectively.

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If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

Mn                         O

% composition                                                      72.1                      27.9

Divide by mass number                                       54.94                     16

1.31                      1.74

Divide by the smallest number                         1.31                      1.31

1                    1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

Consider the following reaction: Br2(g) + Cl2(g) ⇌ 2BrCl(g), Kp=1.112 at 150 K.
A reaction mixture initially contains a Br2 partial pressure of 751 torr and a Cl2 partial pressure of 737 torr at 150 K.
Calculate the equilibrium partial pressure of BrCl.

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g) , Kp=1.112 at 150 K

denoting BC as BrCl , B as Br₂ , C as Cl₂, p as partial pressure , then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √(Kp*pB*pC)

replacing values

pBC = √(Kp*pB*pC) = √(1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr

To calculate the equilibrium partial pressure of BrCl, use the equilibrium constant expression and substitute the given partial pressures of Br2 and Cl2. The equilibrium partial pressure of BrCl is approximately 0.0375 atm.

### Explanation:

To calculate the equilibrium partial pressure of BrCl, we need to use the equilibrium constant expression:

Kp = ([BrCl]^2) / ([Br2] * [Cl2])

Given that the equilibrium partial pressures of Br2 and Cl2 are 0.450 atm and 0.115 atm, respectively, we can substitute these values into the expression:

1.112 = ([BrCl]^2) / (0.450 * 0.115)

Simplifying the expression, we find that the equilibrium partial pressure of BrCl is approximately 0.0375 atm.

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For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.4Al(s) + 302(g) —> 2Al2O3(s)

1 mol Al; 1 mol O2
4 mol Al; 2.5 mol O2
12 mol Al; 10 mol O2
15.4 mol Al; 10.7 mol O2

1 mol Al; 1 mol O2

Explanation:ol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

or the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer aor the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

1 mol Al; 1 mol O2

4 mol Al; 2.5 mol O2

12 mol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

All light travels at the same speed.
O True
O False

False

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