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2 years ago

Calculate the atomic mass of Carbon if the two common isotopes of carbon have masses of

Chemistry

1 answer:

Mars2501 [29]2 years ago

5 0

Answer:

Average atomic mass of carbon = 12.01 amu.

Explanation:

Given data:

Abundance of C¹² = 98.89%

Abundance of C¹³ = 1.11%

Atomic mass of C¹² = 12.000 amu

Atomic mass of C¹³ = 13.003 amu

Average atomic mass = ?

Solution:

Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100

Average atomic mass of carbon= 1186.68 + 14.43333 / 100

Average atomic mass of carbon = 1201.11333 / 100

Average atomic mass of carbon = 12.01 amu.

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sergeinik [125]

The formula for molality---> m = moles solute/ Kg of solvent

the solute here is NH₃ because it's the one with less amount. which makes water the solvent.

1) let's convert the grams of NH₃ to moles using the molar mass

molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol

15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃

2) let's convert the grams of water into kilograms (just divide by 1000)

250.0 g= 0.2500 kg

3) let's plug in the values into the molality formula

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3 years ago

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2 years ago

How many atoms of carbon are in 24 grams of carbon?

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3 years ago

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Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

$CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O$

Number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 2

Number of atoms present on product side are as follows.

C = 1
H = 2
O = 3

To balance this equation, multiply $O_{2}$ by 2 on reactant side. Also, multiply $H_{2}O$ by 2 on product side. Hence, the equation can be rewritten as follows.

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Now, the number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 4

Number of atoms present on product side are as follows.

C = 1
H = 4
O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

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2 years ago