Answer:
The equation for molarity is moles/liter for the first question you would do 0.256/0.143 liters to get 1.790 mol/L
Explanation:
The second problem you would do need to find the moles of NaCl which you would do by doing 4.89 g/58.44g/mol= 0.08367 then do 0.08367/0.600= 0.139 mol/L
The third problem would be the same steps as the second one.
The fourth problem would be (0.460M)(5.50L)= 2.53 moles
CH3COOH
Explanation:
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Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:

Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:

Convert this value to mL:

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
Answer:
0.01M = [H⁺]; 1x10⁻¹²M = [OH⁻]; Ratio is: 1x10¹⁰
Explanation:
pH is defined as -log [H⁺]
For a pH of 2 we can solve [H⁺] as follows:
pH = -log [H⁺]
2 = -log [H⁺]
10^-2 = [H⁺]
<h3>0.01M = [H⁺]</h3>
Using Keq of water:
Keq = 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 0.01M = [OH⁻]
<h3>1x10⁻¹²M = [OH⁻]</h3><h3 />
The ratio is:
[H⁺] / [OH⁻] = 0.01 / 1x10⁻¹² =
<h3>1x10¹⁰</h3>