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3 years ago

8

According to Le Chatelier’s principle, what happens when the concentration of reactants is doubled in a chemical reaction that w

as in equilibrium?

2 answers:

Travka [436]3 years ago

6 0

Answer:

A) The equilibrium shifts to the right to favor the formation of products.

Explanation:

Usimov [2.4K]3 years ago

3 0

If more reactant is added, the equation will shift to the right in order to make more product (which will increase the products)

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Is Benzylamine soluble in water?

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Yes, Benzylamine is miscible, meaning it is soluble at all amounts.

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The [-oh) in a sample of egg whites is 6.3 x 10-7 m. what is the [h3o+] in these<br>egg whites?

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The hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.

<h3>How to calculate [H3O+]?</h3>

The hydrogen ion concentration of a solution can be calculated as follows:

pOH = - log [OH-]

pOH = - log [6.3 × 10-⁷M]

pOH = - [-6.2]

pOH = 6.2

Since pOH + pH = 14

pH = 14 - 6.2

pH = 7.8

pH = - log [H3O+]

7.8 = - log [H3O+]

[H3O+] = 10-⁷:⁸

[H3O+] = 1.5 × 10-⁸M

Therefore, the hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.

Learn more about hydrogen ion concentration at: brainly.com/question/15082545

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"native," or elemental copper can be found in nature, but most copper is mined as oxide or sulfide minerals. chalcopyrite (cufes

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In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

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Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$

where,

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t = time = 10 min = 600 sec (1 min = 60 sec)

Now put all the given values in this formula, we get

$Q=10A\times 600s=6000C$

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = $2\times (1.6\times 10^{-19})$

Number of atoms deposited = $\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}$ atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = $\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113$ moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g

Mass of Copper Deposited = $63.5\times 0.03113 =1.98g$

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