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2 years ago

What are the complete Ionic and Net Ionic equations for the following

Chemistry

1 answer:

zhenek [66]2 years ago

5 0

Answer:

Check photo

Explanation:

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Gwar [14]

Answer:

160.3 g S

Explanation: there you go

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2 years ago

Within the water cycles, where do individual water molecules reside the longest?

solniwko [45]

Condensation and precipitation

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3 years ago

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Plzz help mee asapp dmxkxksms

Korolek [52]

Answer:

1. a part of the soil that contains decayed plant or animal matter

2. erosion

3. topsoil

Explanation:

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3 years ago

Activation energy is: A. The maximum amount of energy reactants can hold. B. The energy needed to begin breaking the bonds of pr

Arte-miy333 [17]

Activation energy is the energy needed to begin breaking the bonds of reactants and is denoted as option D.

<h3>What is a Reactant?</h3>

These are the substances which take part in a chemical reaction and result in the formation of the product.

The activation energy is required as it breaks the bonds of the reactant thereby resulting in the product being formed. This therefore makes it the most appropriate choice.

Read more about Activation energy here brainly.com/question/5280701

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3 0

2 years ago

What is the half-life of 20 g of a radioactive sample if 5 g remain after 8 minutes?

Anni [7]

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) = 20 g

m (final mass after time T) = 5 g

x (number of periods elapsed) = ?

P (Half-life) = ? (in minutes)

T (Elapsed time for sample reduction) = 8 minutes

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

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

$5 = \dfrac{20}{2^x}$

$2^x = \dfrac{20}{5}$

$2^x = 4$

$2^x = 2^2$

$\boxed{x = 2}$

Now, let's find the half-life (P) of the radioactive sample, let's see:

$T = x*P$

$8 = 2*P$

$2\:P = 8$

$P = \dfrac{8}{2}$

$\boxed{\boxed{P = 4\:minutes}}\Longleftarrow(Half-Life)\end{array}}\qquad\checkmark$

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

5 0

2 years ago