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Paladinen [302]
4 years ago
6

Show a numerical setup for calculating the number of moles in the sample of CaCO3

Chemistry
1 answer:
Delvig [45]4 years ago
6 0

The numerical setup is

<span>11<span>g</span></span>CO2×1 mol CO2/44.01g CO2=0.25 mol CO2

gCO2 cancels out so it is

11 x 1 mol CO2/44.1 CO2= 0.25 mol CO2

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blsea [12.9K]

Answer : The half-life at this temperature is, 3.28 s

Explanation :

To calculate the half-life for second order the expression will be:

t_{1/2}=\frac{1}{k\times [A_o]}

When,

t_{1/2} = half-life = ?

[A_o] = initial concentration = 0.45 M

k = rate constant = 6.77\times 10^{-1}M^{-1}s^{-1}

Now put all the given values in the above formula, we get:

t_{1/2}=\frac{1}{6.77\times 10^{-1}M^{-1}s^{-1}\times 0.45M}

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Answer:

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Which option is an isotope with 6 protons and 8 neutrons? (1 point)
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Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.
zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

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The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass =  7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

More about the ideal gas equation link is given below.

brainly.com/question/4147359

#SPJ1

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True, because that’s the right answer lol
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