A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Answer:- 64015 J

Solution: The calorimeter contains 4250 mL of water, which is at a temperature of 22.55 degrees Celsius.

The water's density is 1 gram per mL.

Thus, the mass of water = = 4250 grams.

After introducing the hot copper bar, the final temperature of the water reaches 26.15 degrees Celsius.

Thus, for the water = 26.15 - 22.55 = 3.60 degrees Celsius.

The specific heat capacity of water is 4.184 .

To determine the heat absorbed by the water, we can use the following formula:

where q represents heat energy, m refers to mass, and c indicates specific heat.

Now let's substitute the values into the equation to perform the calculations:

q = 64015 J

Therefore, the water absorbs 64015 J of heat.

The correct answer is: c. N2O, N2O4, N2O5.

According to the law of multiple proportions, also referred to as Dalton's Law, when two elements form compounds, the mass ratios of the second element that combine with a specific mass of the first yield small whole number ratios.

1) For NO, the mass ratio m(N): m(O) is 14: 16, simplified to 7: 8.

2) In N₂O, the ratio m(N): m(O) equates to 2·14: 16, which simplifies to 7: 4.

3) For NO₂, the masses yield m(N): m(O) = 14: 2·16, simplifying to 7: 16.

4) In N₂O₅, the ratio is (2·14): (5·16), which simplifies to 7: 20.

5) For NO₄, the mass ratio is m(N): m(O) = 14: (4·16), which simplifies to 7: 32.

6) N₂O₄ gives a ratio of m(N): m(O) as (2·14): (4·16), simplifying to 7: 16.

A) This means m(N): m(O) = 5.6 g: 3.2 g, simplifying results in 1.75: 1, which further translates to m(N): m(O) = 7: 4.

B) Here, m(N): m(O) is 3.5 g: 8.0 g. Dividing reveals a ratio of 1: 2.285, which alters to m(N): m(O) = 7: 16.

C) Lastly, for m(N): m(O) = 1.4 g: 4.0 g, then adjustments yield a ratio of m(N): m(O) = 7: 20.

To find the answer, start by calculating the total mass of the copper utilized:

Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu

Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:

1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.

Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu

= 4.72 moles CuFeS2

The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole CuFeS2 = 866.49 g CuFeS2