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3 years ago

A 25 mL sample of 0.100 M HNO3 completely reacts with NaOH according to this equation:

2 answers:

3 0

To solve this equation you can use a basic molarity formula and substitute the correct values.

The formula for molarity is M = mol/L.

Start by changing your mL to L. To do this, divide 25 by 1000.
This results in:
.025 L

You can now rearrange your molarity formula to isolate what you are solving for, which is moles.
This leaves your formula as:
ML = mol

Now, plug in your converted volume and given molarity into the formula.
This leaves you with:
(0.100)(0.025) = mol

Multiply to find your final value for moles.
This results in:
0.0025 mol of HNO3

I hope this is what you’re looking for! :)

Angelina_Jolie [31]3 years ago

3 0

Answer:

0.0025

Explanation:

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3 years ago

If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution

Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

$\epsilon$ = molar absorptivity of this solution = $5.56\times 10^4 M^{-1} cm^{-1}$

$0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c$

$c=1.4163\times 10^{-5} M=14.16 \mu M$

( $1\mu M=10^{-6} M$ )

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) $c=1.4163\times 10^{-5} mol/L$

1 L of solution contains $1.4163\times 10^{-5}$ moles of red dye.

Mass of $1.4163\times 10^{-5}$ moles of red dye:

$1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g$

$(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100$

$red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%$

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = $c=1.4163\times 10^{-5} M$

Concentration of red solution after dilution = c'

$c=c'\times 5$

$1.4163\times 10^{-5} M=c'\times 5$

$c'=2.83\times 10^{-6} M$

The final concentration of the diluted solution is $2.83\times 10^{-6} M$

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3 years ago

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Vedmedyk [2.9K]

Answer:

Explanation:

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This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

pH = pKa + log [A⁻]/[HA]

thus

0.1 ≤ [A⁻]/[HA] ≤ 10

Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.

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Answer:

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Symbol: Rb (37)

Atomic Weight: 85.4678

Atomic Number: 37

Number of Stable Isotopes: 1 (View all isotope .

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