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3 years ago

1. How many moles of carbon are in 2.25 * 10^22 atoms of carbon ?

Chemistry

1 answer:

tresset_1 [31]3 years ago

3 0

Answer:

atoms of C? 2.25 x 1022 atoms C x 1 mole C = 0.037 mol C

Explanation:

How many moles of C are in 2.25 x 1022 atoms of C?

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Answer: 0.05

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What do u need to know to describe the velocity of an object?

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Consider the reaction of CaCN2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How m

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Answer:

56 g. Option 3.

Explanation:

The reaction is: CaCN₂ + 3H₂O → CaCO₃ + 2 NH₃

1 mol of calcium cianide reacts with 3 moles of water in order to produce 1 mol of calcium carbonate and 2 moles of ammonia

We have the mass of each reactant, so let's convert the mass to moles:

45 g. 1mol / 80.08 g = 0.562 moles of cianide

45 g. 1mol / 18 g = 2.5 moles of water

The cianide is the limiting reactant:

3 moles of water need 1 mol of cianide to react

Then, 2.5 moles of water will need (2.5 . 1)/ 3 = 0.833 moles

As we have 0.562 moles of CN⁻ we don't have enough

We can work now, on the reaction:

Ratio is 1:1. Therefore 0.562 moles of cianide will produce 0.562 moles of carbonate

Let's convert the mass to moles to find the answer:

0.562 mol . 100.08 g / 1 mol = 56.2 g

8 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 $91^{o}C$ .

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3 years ago

What compound was formed by the reaction of the first oxygen released by cyanobacteria and iron?

castortr0y [4]

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4 years ago