to form brass, copper and zinc combine but do not chemically bond. brass is a(n) ________. a atom b element c compound d

Chemistry

1 answer:

umka21 [38]3 years ago

3 0

Answer: Mixture

Explanation:To form brass, copper and zinc combine but do not chemically bond. Brass is a(n)

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15. Which sample of argon gas has the same number of atoms as a 100.-milliliter sample of helium gas at 1.0 atm and 300. K? A) 5

OLga [1]

The sample of argon gas that has the same number of atoms as a 100 milliliter sample of helium gas at 1.0 atm and 300 is 100. mL at 1.0 atm and 300. K

The correct option is D.

<h3>What is the number of moles of gases in the given samples?</h3>

The number of moles of gases in each of the given samples of gas is found below using the ideal gas equation.

The ideal gas equation is: PV/RT = n

where;

P is pressure
V is volume
n is number of moles of gas
T is temperature of gas
R is molar gas constant = 0.082 atm.L/mol/K

Moles of gas in the given helium gas sample:

P = 1.0 atm, V = 100 mL or 0.1 L, T = 300 K

n = 1 * 0.1 / 0.082 * 300

n = 0.00406 moles

For the argon gas sample:

A. n = 1 * 0.05 / 0.082 * 300

n = 0.00203 moles

B. n = 0.5 * 0.05 / 0.082 * 300

n = 0.00102 moles

C. n = 0.5 * 0.1 / 0.082 * 300

n = 0.00203 moles

D. n = 1 * 0.1 / 0.082 * 300

n = 0.00406 moles

Learn more about ideal gas equation at: brainly.com/question/24236411

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1 year ago

1. A gas having the following composition is burnt under a boiler with 50% excess air.

jeka94

The composition of the stack gas are :

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of $CH_4$ : 70% of 100 g = 70 g

Mass of $C_3H_8$ : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of $CH_4$ : $\frac{70 g}{16.04 g/mol}$ = 4.3 mole