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2 years ago

How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?

Chemistry

1 answer:

saw5 [17]2 years ago

5 0

Answer:

10437calories

Explanation:

The following data were obtained from the question given:

M = 347.9g

C = 4.2J/g°C

T1 = 25°C

T2 = 55°C

ΔT = 55 — 25 = 30°C

Q =?

Q = MCΔT

Q = 347.9 x 4.2 x 30

Q = 43835.4J

Converting this to calories, we obtained the following:

4.2J = 1 calorie

43835.4J = 43835.4/ 4.2 = 10437calories

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Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel

n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

Answer: The mole fraction of gas B (hydrogen gas) is 0.557

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For nitrogen gas:

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol$

For hydrogen gas:

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol$

For ammonia gas:

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B+n_C}$

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

$\chi_{(H_2)}=\frac{1.025}{1.84}=0.557$

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

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3 years ago

Which of the following is true about compounds

Anastasy [175]

Compounds are composed of 2 or more different elements, ionically and or covalently bonded that results in substances containing elements in their respective definite amounts within the compound.

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3 years ago

Is hydrogen gas a mixture or pure substance?

Akimi4 [234]

Answer:

pure hydrogen is a pure substance even though it consists of many different types of molecules. what makes it pure substance is that it is free from contamination.

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2 years ago

Does anyone know this

LekaFEV [45]

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Your answer would be (D

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2 years ago

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There are 100.0 grams of each reactant available determine the limiting reactant in this equation

Romashka [77]

Since you have not included the chemical reaction I will explain you in detail.

1) To determine the limiting agent you need two things:

- the balanced chemical equation

- the amount of every reactant involved as per the chemical equation

2) The work is:

- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.

- determine the number of moles of each reactant with this formula:

number of moles = (mass in grams) / (molar mass)

- set the proportion with the two ratios (theoretical moles and actual moles)

- compare which reactant is below than the stated by the theoretical ratio.

3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:

i) Chemical equation: H₂ + O₂ → H₂O

ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O

iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂

iv) Covert 100 g of H₂ into number of moles

n = 100g / 2g/mol = 50 mol of H₂

v) Convert 100 g of O₂ to moles:

n = 100 g / 32 g/mol = 3.125 mol

vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂

vii) Compare the two ratios:

2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂

Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.

Therefore, the O₂ is the limiting reactant in this example.

7 0

2 years ago