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zhenek [66]
3 years ago
6

What is the ph of a solution with hydrogen ion concentration of 0.001m?

Chemistry
1 answer:
lys-0071 [83]3 years ago
6 0
Find the pH using the concentration of hydrogen ions [H+] of 0.001M (molarity) by taking the negative log of the value:

-log[H+]=pH
-log[0.001] = pH
-log[1x10^-3]=pH
3=pH
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The half-life of 3H (H-3) is 12 years. About how long does it take for 127/128 of a sample of that radionuclide to decay? (Hint:
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Table salt is the compound sodium chloride (NaCl). Use the periodic table to find the molar mass of NaCl
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6 0
3 years ago
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Question 4
S_A_V [24]

Answer:

Heat is released.

Explanation:

In an exothermic reaction, the reactant(s) move into a more stable state when they form into the product(s). Thus, the excess energy is released in the form of heat.

Conversely, in an endothermic reaction, the reactant(s) move into a more unstable state. In doing so, energy is required for the chemical reaction to occur to force the reactant(s) into a less stable state. So, energy is absorbed to make the reaction occur.

Also, since energy was released, the net energy decreased. If energy was absorbed in an endothermic reaction, the net energy would increase.

Therefore, the answer is 'Heat is released'.

4 0
3 years ago
Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces
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Answer:  The concentration of excess [OH^-] in solution is 0.017 M.

Explanation:

1. Molarity=\frac{moles}{\text {Volume in L}}

moles of HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles

1 mole of HCl give = 1 mole of H^+

Thus 0.019 moles of HCl give = 0.019 mole of H^+

2. moles of Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles

According to stoichiometry:

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Thus 0.012 moles of Ba(OH)_2 give = 2 \times 0.012=0.024 moles of OH^-

H^++OH^-\rightarrow H_2O

As 1 mole of H^+ neutralize 1 mole of OH^-

0.019 mole of H^+ will neutralize 0.019 mole of OH^-

Thus (0.024-0.019)= 0.005 moles of OH^- will be left.

[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M

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4 0
3 years ago
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