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neonofarm [45]
3 years ago
8

Assuming the final solution will be diluted to 1.00 l , how much more hcl should you add to achieve the desired ph?

Chemistry
1 answer:
solong [7]3 years ago
4 0
H = 2.7, therefore amount of H+ needed is 10^-2.7 M 

<span>u haf 80 ml of hcl and 90 ml of naoh left, therefore 20 ml of hcl used, and 10ml of naoh used. </span>

<span>mols of H+ = 0.02 x 7 x 10^-2 = 1.4 x 10^-3 </span>
<span>mols of OH- = 0.01 x 5 x 10^-2 = 5 x 10^-4 </span>

<span>H+ and OH- neutralise each other, so remaining mols of H+ = 9 x 10^-4 </span>

<span>u need 10^-2.7 mols of H+, so 10^-2.7 - 9 x 10^-4 = 0.001095 mol </span>

<span>vol of HCL needed = 0.001095 / 7 x 10^-2 = 0.0156 L = 15.6 mL</span>
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Answer:

H₂: 0.48,  N₂: 0.43;  Ar: 0.09

Explanation:

First of all, sum all the pressures to know the total pressure in the mixture.

434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr

Mole fraction = Pressure gas / Total Pressure

Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48

Mole Fraction N₂: 389.9 /901.8 Torr =0.43

Mole Fraction Ar:  77.9 /901.8 Torr = 0.09

Remember: <u>SUM OF MOLE FRACTION = 1</u>

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Which of the following is a variable property for a gas ? Select all that apply.
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A student uses a sample of KOH stock solution and dilutes it to a total of 120 mL. If the diluted solution is 0.60 M KOH and its
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What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen
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Answer: Mg_{1}Cl_{2}O_{8}

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg=\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles

Moles of Cl = \frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles

Moles of O =\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = \frac {0.45}{0.45}=1

For Cl = \frac {0.89}{0.45}=2

For O= \frac {3.58}{0.45}=8

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is Mg_{1}Cl_{2}O_{8}

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