All categories

3 years ago

How many moles are in 29.5 grams of Ax?

Chemistry

1 answer:

JulsSmile [24]3 years ago

6 0

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

Ar = 39.95 g/mole

The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

$= \frac{29.5}{39.95} \\\\= 0.74 \ mole$

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.

<em>"Your question seems to be missing the correct symbol for the element" </em>

Argon = Ar

Learn more here:brainly.com/question/4628363

You might be interested in

Help me please I really need help on this?

vovangra [49]

Explanation:

first one d

second cant see clear

3 0

3 years ago

Find the molecules in 32 grams of CH4

Elena L [17]

════════ ∘◦❁◦∘ ════════

<h3>Answer = 2</h3>

════════════════════

<h3>Known</h3>

Mass = 32grams

Name of atom = Methane (CH4)

Molar mass C = 12

Molar mass H = 1

════════════════════

<h3>Question</h3>

molecules (mol?)

════════════════════

mass = mol × molar mass

32g = mol × (CH4)

33g = mol × (12 + (4×1))

32g = mol × 16

mol = 32 : 16

mol = 2

════════════════════

#Ey tell me if its mol or total particles

8 0

3 years ago

____ 26. what is the weight percentage of nitrogen in urea, cn2h4o

Paraphin [41]

Weight percentage of nitrogen can be calculated using the following rule:
weight percentage of nitrogen = (weight of nitrogen / weight of urea) x 100

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of hydrogen = 1 grams
molecular mass of oxygen = 16 grams

therefore:
mass of nitrogen in urea = 2(14) = 28 grams
mass of urea = 12 + 2(14) + 4(1) + 16 = 60 grams

Substitute with the masses in the equation to get the percentage:
weight percentage of nitrogen = (28/60) x 100 = 46.667%

7 0

4 years ago

Prove that PV = nRT.

qaws [65]

Find your answer in the explanation below.

Explanation:

PV = nRT is called the ideal gas equation and its a combination of 3 laws; Charles' law, Boyle's law and Avogadro's law.

According to Boyle's law, at constant temperature, the volume of a gas is inversely proportional to the pressure. i.e V = 1/P

From, Charles' law, we have that volume is directly proportional to the absolute temperature of the gas at constant pressure. i.e V = T

Avogadro's law finally states that equal volume of all gases at the same temperature and pressure contain the same number of molecules. i.e V = n

Combining the 3 Laws together i.e equating volume in all 3 laws, we have

V = nT/P,

V = constant nT/P

(constant = general gas constant = R)

V = RnT/P

by bringing P to the LHS, we have,

PV = nRT.

Q.E.D

6 0

3 years ago

Ammonia (NH3) is one of the most common chemicals produced in the united states. it is used to make fertilizer and other product

Greeley [361]

Answer:

a) N2 is the limiting reactant

b) <u>1215.9 grams NH3</u>

<u>c) 2283.6 grams H2</u>

Explanation:

Step 1: Data given

Mass of N2 = 1000 grams

Molar mass N2 = 28 g/mol

Mass of H2 = 2500 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) +3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 1000 grams / 28.0 g/mol

Moles N2 = 35.7 moles

Step 4: Calculate moles H2

Moles H2 = 2500 grams / 2.02 g/mol

Moles H2 = 1237.6 moles

Step 5: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

<u>N2 is the limiting reactant</u>. There will be consumed 35.7 moles.

H2 is in excess. There will react 3*35.7 = 107.1 moles

There will remain 1237.6 - 107.1= 1130.5 moles H2

This is 1130.5 * 2.2 = <u>2283.6 grams H2</u>

Step 6: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 35.7 moles N2 we'll have 2*35.7 = 71.4 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 71.4 moles * 17.03 g/mol

Mass NH3 = <u>1215.9 grams NH3</u>

7 0

3 years ago