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3 years ago

Calculate the number of ATOMS in 1.0 mole of O2.

Chemistry

2 answers:

podryga [215]3 years ago

8 0

Answer: The number of atoms in given amount of oxygen gas is $6.022\times 10^{23}$

Explanation:

According to the mole concept:

1 mole of a substance contains $6.022\times 10^{23}$ number of atoms.

We are given:

Moles of oxygen gas = 1 mole

So, the number of atoms that will be present in 1 mole of oxygen gas will be $1\times 6.022\times 10^{23}=6.022\times 10^{23}$ number of atoms.

Hence, the number of atoms in given amount of oxygen gas is $6.022\times 10^{23}$

Wewaii [24]3 years ago

7 0

There is 6.02*10^23 molecule per mole. And there is 2 atoms per oxygen molecule. So the answer is 1.204*10^24 atoms in 1.0 mole of O2.

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How are cells the basic units to all organisms??

Nonamiya [84]

Answer:

Cells as Building Blocks

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3 years ago

An oxide and carbonate containing oxygen

exis [7]

Answer:

example of carbonate containing oxygen :

CaCO3

( calcium carbonate having 3 atoms of oxygen )

example of oxide containing oxygen :

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5 0

3 years ago

I need help with part B . please

myrzilka [38]

So it's good to map out what you know you have and work from there:
We have two liter measurements and one mole measurement, and we need to find the moles.

For this problem, think of it this way: 46 liters of gas = 1.4 moles.
If one side changes, the other has to as well (if the liters decrease, the moles decrease. if the liters increase, so do the moles.) What you can do is put this into a fraction:

 1.4 moles
 46 L 

if we know that each liter of gas is equal to x amount of moles, we know that 11.5 liters equals some amount of moles. You can put this into a fraction too, and make it equal to the other fraction:

1.4 moles = x moles
46 L 11.5 L

Then get your calculator out and do some algebra.

11.5 * (1.4/46) = x

The answer should come out to be: 0.35 moles

8 0

3 years ago

How many grams of calcium chloride are needed to produce 1.50 g of potassium chloride? cacl2(aq) + k2co3(aq) → 2 kcl(aq) + caco3

Virty [35]

CaCl2(aq) + K2CO3(aq) → 2 KCl(aq) + CaCO3(aq)

1.12 g
2.23 g
0.896 g
4.47 g
1.12 g

Hope it helps :)

8 0

3 years ago

Read 2 more answers

Potassium nitrate decomposes on heating, producing potassium oxide and gaseous nitrogen and oxygen:

ahrayia [7]

To produce 56.6 kg of oxygen, moles of KNO₃ = 142.915 kg

This is a stoichiometric kind of question where everything should be converted into moles in order to proceed on with the calculations.

First you find the number of moles of oxygen using the following formula:

Mole (mol) = Mass (g) / Molar Mass (g/mol)

For oxygen, the molar mass is 32g/mol.

Therefore the number of moles in 56.6 kg of oxygen is :

(56.6 × 1000)/32

= 1768.75

From the reaction equation, the ratio of $KNO_{3}$ to O2 is 4:5.

Thus the number of moles of $KNO_{3}$ is

1768.75 × 4/5

= 1415 mol.

If you are looking for the mass of $KNO_{3}$ needed then Mass (g)

= Mole (mol) × Molar Mass (g/mol),

1415 ×101 = 142,915g

= 142.915kg

What happens to potassium nitrate when heated?

When potassium nitrate is heated, it decomposes into potassium nitrite and oxygen. Write a balanced equation for this reaction and add the state symbols of the reactants and products.

Learn more about molar mass :

brainly.com/question/24236159

#SPJ4

4 0

2 years ago