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nata0808 [166]
3 years ago
5

How does activation energy affect chemical reactions?

Chemistry
2 answers:
Assoli18 [71]3 years ago
8 0

Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.

<em>Best of luck,</em>

<em>-Squeak</em>

valentinak56 [21]3 years ago
4 0

Answer:

A lower activation energy means more rate of reaction.

Explanation:

The activation energy is the extra minimum amount of energy required by reactant molecules to undergo a chemical reaction.

All reactant molecules have an inherent internal energy. They need to cross an energy barrier known as activation energy, to undergo conversion to product. The energy is required for effective collision between reactant molecules.

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Answer:

B. a nuclear reactor core (1000°C)......

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1. 48 mols

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Explanation:

Molarity= mols divided by liters

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3 years ago
Weight of one mole of carbon = 12.01 g Weight of one mole of oxygen = 16.00 g The molecular weight (gram formula weight) for CO
murzikaleks [220]

Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

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equal to the sum of the weight of both elements.

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4 years ago
Write and label the density triangle
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3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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