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Ostrovityanka [42]
3 years ago
15

The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample changed

by 20.0°C, how many calories of heat energy were absorbed by the sample?
6.2 calories
0.016 calories
6,451 calories
0.062 calories
Chemistry
2 answers:
kramer3 years ago
6 0
I believe the correct answer here is 6.2. Since you would have to multiply the specific heat of gold by the amount of gold by the new heat of the gold. So this would be:

0.031 x 10 x 20 = 6.2
eimsori [14]3 years ago
3 0
The answer is A. 6.2 calories
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what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg
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Answer:

The molar mass is 180.2 g/mol

Explanation:

<u>Step 1:</u> Data given

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Temperature = 20 °C

Pressure = 1 atm

<u>Step 2:</u> Calculate the number of moles

PV = nRT

 ⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

<u>Step 3</u>: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

<u>Step 4:</u> Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F =  3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

<u>Step 5:</u> Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0
3 years ago
Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of carbon tetrachlori
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Answer:

0.55mL of carbon tetrachloride

Explanation:

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From the balanced reaction equation

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If 1.1mL of chlorine were consumed, volume of carbon tetrachloride= 1.1×22400/44800

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Note: 1 mole of a gas occupies 22.4L volume or 22400mL

7 0
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