If 200 mL of a 2.50 M #NaOH# solution is diluted to 500 mL, what is the new concentration of NaOH?
The new concentration is
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The new concentration of NaOH after diluting 200 mL of a 2.50 M solution to 500 mL can be calculated using the formula for dilution:
[C_1V_1 = C_2V_2]
Where: (C_1) = initial concentration (V_1) = initial volume (C_2) = final concentration (V_2) = final volume
Substituting the given values:
(C_1 = 2.50 , \text{M}) (V_1 = 200 , \text{mL}) (V_2 = 500 , \text{mL})
[2.50 , \text{M} \times 200 , \text{mL} = C_2 \times 500 , \text{mL}]
[500 = C_2 \times 500]
[C_2 = \frac{500}{500}]
[C_2 = 1.00 , \text{M}]
So, the new concentration of NaOH after dilution is 1.00 M.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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