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2 years ago

What is the ph of a solution with a hydroxide ion, oh−, concentration of 5.4 x 10−4 m?

1 answer:

7 0

POH = -log[OH-]
pH + pOH = 14 (at 25 degrees celsius)

pOH = -log(5.4*10^-4) = 3.27

pH = 14 - 3.27 = 10.73

The pH of a solution with a hydroxide ion concentration of 5.4*10^-4 M is 10.73.

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When an atom of the unstable isotope Na-24 decays, it becomes an atom of Mg-24 because the Na-24 atom spontaneously releases:(1)

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When an atom of the unstable isotope Na-24 decays, it becomes an atom of Mg-24 because the Na-24 atom spontaneously releases (2) a<span> beta particle</span>.

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2 years ago

Polymerization starts at the ________ end of the template.

balandron [24]

Answer:

B. 3'

Explanation:

Polymerization is the process of forming large molecules by joining together many small molecules. The small molecules are known as the monomers, and the macromolecules are called polymers.

There are two(2) types of polymerization.

a. addition polymerization

b. condensation polymerization

Polymerization is characterized by eliminating small molecules such as water, carbon(IV)oxide, methanol e.t.c.

In complex compounds, polymerization starts at the 3' end of the template.

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3 years ago

A balloon is filled with 3.00 L of helium at a pressure of 765 torr. What is the volume of the balloon at an altitude where the

Sidana [21]

Answer:

4.33L

Explanation:

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2 years ago

Describe the relationship between pure chemistry and applied chemistry

True [87]

Pure chemistry: gain knowledge for own pleasure
Applied chemistry: gain knowledge to know how to use it

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2 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago