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yulyashka [42]
2 years ago
13

CH3CH3Li ionic or covalent

Chemistry
1 answer:
vampirchik [111]2 years ago
4 0

Answer: CH3CH3Li is in fact Ionic!

Explanation:

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Artyom0805 [142]

Answer:

A

Explanation:

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3 years ago
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How many atp molecules are produced during the citric acid cycle?
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"Only 2 molecules" of ATP <span>produced during the citric acid cycle

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At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the
Elis [28]

Answer:

P₂  = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂ 

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂  =  25.83 atm .L.  K  / 293 K . L

P₂  = 0.09 atm

7 0
3 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
2 years ago
1/4 checkpoint
DochEvi [55]

Answer:

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Explanation:

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