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2 years ago

. If the combustion of 45.0 g of methane (natural gas) releases 2498 kJ of heat energy, how much heat

Chemistry

2 answers:

Amanda [17]2 years ago

6 0

Answer:

8.60 x 10⁴ kJ

Explanation:

The energy released per gram of methane is first calculated:

(2498kJ)/(45.0g) = 55.5 kJ/g

The heat released by the combustion of 1.55 kg of methane is then calculated:

(55.5kJ/g)(1.55kg)(1000g/kg) = 86025kJ

To three significant digits, the answer is 8.60 x 10⁴ kJ.

blondinia [14]2 years ago

5 0

45g ---> 2498kJ
1.55g ----> ?kJ
Answer: (1.55x2498)/45 approx 86.04 (kJ)

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Zn + 2 HCl --> ZnCl2 + H2 How many moles of hydrochloric acid (HCl) are used (assume a complete reaction) if 8.3 moles of zin

uysha [10]

Answer:

E. 16.6 mol HCl

Explanation:

The equation for the reaction is;

Zn + 2 HCl --> ZnCl2 + H2

From the reaction 2 moles of HCl produces 1 mole of ZnCl2

Therefore; 8.3 moles of ZnCl2 will be produced by;

= 8.3 moles ×2

= 16.6 Moles of HCl

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2 years ago

A balloon containing 0.0400 mol of a gas with a volume of 500 mL was expanded to 1.00 L.

ANTONII [103]

Answer:

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2 years ago

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salantis [7]

Answer:

The increase in the atomic number

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2 years ago

Which of the following aqueous solutions has the highest boiling point?A. 1.25 M Ca(NO3)2B. 1.25 M KNO3C. 1.25 M CH3OHD. 2.50 M

Drupady [299]

Answer:

Ca(NO3)2 has the highest boiling point ( option A)

Explanation:

Step 1: Data given

A. 1.25 M Ca(NO3)2

B. 1.25 M KNO3

C. 1.25 M CH3OH

D. 2.50 M C6H12O6

Step 2: Calculate highest boiling point

The boiling point depends on the van't Hoff factor

This shows the particles produced when the substance is dissolved. For non-electrolytes dissolved in water, the van' t Hoff factor is 1.

Ca(NO3)2 → Ca^2+ + 2NO3- → Van't Hoff factor = 3

KNO3 → K+ + NO3- → Van't Hoff factor = 2

CH3OH is a non-elektrolyte → Van't Hoff factor = 1

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2 years ago

Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0g of potassium and

Leona [35]

Answer:

79.43kg

Explanation:

To adequately solve this problem, we should know the law of constant composition. This is a pointer to the fact that no matter the type of sample, the percentage compositions of potassium and iodine still remains the same. We can use these masses to get a formula and we would know the percentage compositions in whatever mass we are dealing with.

First of all, we add the masses in the first sample. 13 + 42.3 = 55.3

Hence, the percentage composition of the potassium is 13/55.3 * 100 = 23.51%

The percentage composition of the iodine is = 100 - 23.51 = 76.49%

Now, we need to get the formula of the compound. We can get this by dividing the percentage compositions with the atomic masses. The atomic mass of potassium and iodine is 39 and 127 respectively.

Potassium = 23.51/39 =0.603

Iodine = 76.49/127 = 0.602

We then divide by the smaller value to get the formula and this shoes our formula is KI

We can see they have a ratio of 1 to 1, meaning one atom of potassium to one atom of iodine. This further confirms the percentage compositions of 23.5 to 76.5

Now to get the mass of iodine yielded, let us say the mass is xkg

This means x/(x + 24.4) * 100 = 76.5

100x = 76.5( x + 24.4)

100x = 76.5x + 1866.6

100x - 76.5x = 1866.6

23.5x = 1866.6

x = 1866.6/23.5 = 79.43kg

4 0

2 years ago