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Kryger [21]
3 years ago
12

What will be the mass of carbon dioxide

Chemistry
1 answer:
Elza [17]3 years ago
4 0
44.01 g/mol is the mass
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Caed for this question.
OLga [1]

Answer:

0.962 atm.

97.4 kPa.

731 torr.

14.1 psi.

97,434.6 Pa.

Explanation:

Hello.

In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:

- Atmospheres: 1 atm = 760 mmHg:

p=731mmHg*\frac{1atm}{760mmHg} =0.962atm

- Kilopascals:: 101.3 kPa = 760 mmHg:

p=731mmHg*\frac{101.3kPa}{760mmHg} =97.4kPa

- Torrs: 760 torr = 760 mmHg:

p=731mmHg*\frac{760 torr}{760mmHg} =731 torr

- Pounds per square inch: 14.69 psi = 760 mmHg:

p=731mmHg*\frac{14.69}{760mmHg} =14.1psi

- Pascals: 101300 Pa = 760 mmHg:

p=731mmHg*\frac{101300Pa}{760mmHg} \\\\p=97,434.6Pa

Best regards.

5 0
3 years ago
Calculate the volume of the gas, in liters, if 1.75 mol has a pressure of 1.28 atm at a temperature of -7 ∘C
Alexandra [31]

Answer:

A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.

 

You need to apply the ideal gas law PV=nRT

 

You have the pressure, P=1.01 atm

you have the volume, V = 2.21 L

The ideal gas constant R= 0.08205 L. atm/ mole.K at  273 K

 

find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)

 

n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and

L=1.23 L

 

P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L

= 1.994 atm

Explanation:

6 0
3 years ago
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 3.50 moles of magnesium
Firdavs [7]

Answer:

n_{Mg}=3.50molMg\\\\ n_{Cl}=7.00molCl\\\\n_O=28.0molO

Explanation:

Hello there!

In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

n_{Mg}=3.50molMg(ClO_4)_2*\frac{1molMg}{1molMg(ClO_4)_2}=3.50molMg\\\\ n_{Cl}=3.50molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2}=7.00molCl\\\\n_O=3.50molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2}=28.0molO

Best regards!

3 0
3 years ago
What is the name of this molecular compound ? F6I5 PLEASE AND THANK YOU
lawyer [7]

Answer:

Hexaflourine Pentaiodide?

Explanation:

f = flourine (6 = hexa)

i = iodine (5 = penta) + ide

5 0
3 years ago
Calculate the theoretical yield for the bromination of both stilbenes
podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0
2 years ago
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