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Sloan [31]
3 years ago
10

How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?

Chemistry
1 answer:
PIT_PIT [208]3 years ago
7 0
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
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In column chromatography (microscale), after loading it with solvent and adsorbent and prior to loading the sample, what level s
Nikolay [14]

Answer:

Explanation:

You should allow the solvent to drop to the level of the adsorvent, so it would never run dry.

When you let your sample to run dry it will never finish to flow from the adsorbent depending of it polarity.

Water should not be used because it can dissolve the adsorbent.

You could use another technique to identify the compound, as an infrared or a ultraviolet detector. You can also, if you know the compounds, identify it for the retention time, for example, if you need to detect two compounds, one more polar than the other, and use a polar adsorbent and a non-polar solvent, the first compound to exit the column will be the less polar one, because it will have a bigger interaction with the solvent than the stationary phase (adsorbent) and will go faster, the second will be the more polar one, because it will have a bigger interaction with the stationary phase.

5 0
3 years ago
How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?
statuscvo [17]

Answer:

A) 6.5\times 10^{23}\ \text{molecules}

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

The reaction of electrolysis would be

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)

Number of moles of H_2O

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}

From the reaction it can be seen that 2 moles of H_2O gives 1 mole of O_2

So, number of moles of O_2 produced is

\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}

Number of molecules

1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}

So, 6.5\times 10^{23}\ \text{molecules} of oxygen is produced.

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2 years ago
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5 0
3 years ago
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A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t
svp [43]
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>

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<span>Then convert to kJ/mol


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3 years ago
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lisov135 [29]
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