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elena-s [515]
3 years ago
12

HELP PLEASE ITS JUST MULTIPLE CHOICE

Chemistry
1 answer:
morpeh [17]3 years ago
6 0

Answer:

Atoms are indivisible

I had a hard time finding this out.

Explanation: Dalton's atomic theory was the first complete attempt to describe all matter in terms of atoms and their properties. Dalton based his theory on the law of conservation of mass and the law of constant composition. The first part of his theory states that all matter is made of atoms, which are indivisible.

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HELP ME ON THIS
diamong [38]

Answer:

The attractive force between them decreases

Explanation:

This is because they become localised.

7 0
2 years ago
7. Diethyl ether burns in air according to the following equation. C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l) If 7.15 L of CO2 is
iren [92.7K]

From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.

<h3>What is combustion?</h3>

Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)

We can obtain the number of moles of CO2 from;

PV = nRT

n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K

n = 7.29 /32.6

n = 0.22 moles

If 6 moles of oxygen produces 4 moles of CO2

x moles of oxygen produces 0.22 moles of CO2

x = 0.33  moles

1 mole of oxygen occupies 22.4 L

0.33 moles of oxygen occupies 0.33 moles *  22.4 L/ 1 mole

= 7.4 L of oxygen

Learn more about stoichiometry: brainly.com/question/13110055

#SPJ1

6 0
2 years ago
Which order is correct in listing the bond lengths from shortest to longest? A. single, double, triple B. triple, double, single
Alex787 [66]
A I believe would be the answer
3 0
3 years ago
PLEASE ANSWER QUICKLY I'M IN A TEST
Natasha_Volkova [10]

D is the correct answer......

hope it help, please thank me if it did.....

4 0
3 years ago
Read 2 more answers
How to balance this equation KClO4 → KCl + ?O2(g) and what type of reaction occurs?
Ivahew [28]

Explanation:

here is the answer to your question

7 0
3 years ago
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