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3 years ago

When you strike a match, what do you provide that enables the flame to appear?

Chemistry

1 answer:

Alexeev081 [22]3 years ago

3 0

Explanation:

The minimum amount of energy required by a substance to begin a reaction is known as the activation energy.

Therefore, we can conclude that when you strike a match, you provide activation energy that enables the flame to appear.

Whereas a catalyst helps in increasing the rate of reaction but it does not start the reaction. On the other hand, when reactants combine together then we get products.

Thus, we can conclude that options (a), (b), (c), and (e) are not correct.

Hence, option (d) is the correct answer.

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The scientific method is a system of reasoning. You can make reasonable explanations for something that you are observing in the world.

Explanation:

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3 years ago

Which element has the same number of energy levels as aluminum (Al) and the same number of valence electrons as calcium (Ca)

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3 years ago

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Which statement is true about the patterns shown on the electromagnetic spectrum?

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A. As wavelength increases, radiant energy and frequency increase.

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2 years ago

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0

3 years ago

HELP! ASAP! Iron (Fe) and copper (II) chloride (CuCl2) combine to form iron (III) chloride (FeCl3) and copper (Cu). If you start

Helga [31]

Answer: 30.978

Explanation:

From the equation 2 moles of Fe will result in 3 moles copper

so .325 moles Fe will result in .4875 moles Cu

Cu weights 63.546 gm per mole

.4875 moles * 63.546 gm / mole = 30.978 gm of Cu

7 0

2 years ago