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lisov135 [29]
4 years ago
8

1. How many ATOMS of carbon are present in 7.48 grams of carbon monoxide ?

Chemistry
1 answer:
Salsk061 [2.6K]4 years ago
3 0

Answer:

The answer is 1.61 × 10²³ atoms

Explanation:

To determine number of atoms, we will use the formula below

Number of atoms = number of moles (n) × avogadro's constant (6.02 x 10²³)

n was not provided, hence we will solve for n

n = mass/ molar mass

molar mass of carbon monoxide, CO (where C is 12 and O is 16) is 12 + 16 = 28

mass was provided in the question as 7.48

n = 7.48/28

n = 0.267

Hence,

number of atoms = 0.267 × 6.02 x 10²³

= 1.61 × 10²³ atoms

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Calculate   the  number  of  each   reactant  and  the  moles  of  the  product
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The  moles  of  CaO=  56.08g/  56.08g/mol(molar  mass  of  Cao)=  1mole
the  moles  of  water=  36.04 g/18  g/mol=  2.002moles
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 The  mass  of differences  of  reactant  and  product  can   be  therefore 
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3 years ago
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Answer:

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<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

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Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

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