Answer to first question

The summary or ending of your experiment

Katyanochek1 [597]

Answer:

Conclusion

Explanation:

I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.

7 0

4 years ago

A 0.100 mile sample of gas is at a temperature of 85.0 degrees C and a volume of 3.47 L. What is the pressure of the gas( in mm

vladimir1956 [14]

Answer: 64.6 mmHg

Explanation:

Given that:

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

Recall that pressure of the gas is required in mm hg, so convert 0.085 atm to mm Hg

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

Thus, the pressure of the gas is 64.6 mm hg

6 0

3 years ago

when 0.72 g of a liquid is vaporized at 110° C and 0.967 atm, the gas occupies a volume of 0.559L. The empirical formula of the

Leno4ka [110]

Hydrogen gas mixed with sulfur

8 0

3 years ago

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

My name is Ann [436]

Answer:

602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The yield is measured by the amount of each product produced by the reaction.

The chemical formula of fluorene is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The oxidation, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

$2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

1. Mole ratio

$2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

2. Convert 175mg of fluorene to number of moles

$175mg/times 1g/1,000mg=0.175g$

Number of moles = mass in grams / molar mass

$\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

3. Set a proportion for each product of the reaction

a) For CO₂

i) number of moles

$2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

mass = number of moles × molar mass

mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) For H₂O

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

mass = number of moles × molar mass

mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg

4 0

3 years ago

what mass of carbon dioxide will be produced when 12.9 g of butane reacts with an excess of oxygen in the following reaction?

Nadya [2.5K]

39.1 gCO2
I think if that’s an option

3 0

3 years ago