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Ulleksa [173]
3 years ago
6

How many grams of a stock solution that is 87.5 percent H2SO4 by mass would be needed to make 275 grams of a 55.0 percent by mas

s solution? Show all of the work needed to solve this problem.
Chemistry
2 answers:
cricket20 [7]3 years ago
7 0
Take a look into this beautiful part and you will get your answer
<span>275 grams of a 55 per cent solution contains 0.55* 275 grams of pure H2SO4. We need to find the mass of 87.5 per cent solution which contains 0.55*275 grams H2SO4 this is (0.55* 275) / 0.875 grams. In that way you will have your answer. I hope this can help</span>
levacccp [35]3 years ago
7 0

The mass of sulfuric acid in the required solution is 55g per 100g of solution

We need to make total of 275 grams of solution of concentration 55%

So we need following amount of sulfuric acid in the solution

55% of 275  grams = 0.55 X 275 = 151.25 grams

Now we have a stock solution of sulfuric acid 87.5% (w/w)

It means in every 100g of solution 87.5g of sulfuric acid

We need 151.25 grams of sulfuric acid

87.5g of sulfuric acid will be obtained from 100g of stock solution

1g of sulfuric acid will be obtained from 100/87.5g of stock solution

151.25g of sulfuric acid will be obtained from = \frac{100X151.25}{87.5}=172.85g

Hence we will take 172.85 grams of stock solution to prepare 55% sulfuric acid solution of mass 275g

Rest we will take water.

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How do polar and nonpolar covalent bonds differ?
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Answer: In polar covalent bonds a pair of electrons are unequally shared between two atoms, while in nonpolar covalent bonds two atoms share a pair of electrons with each other.

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Explanation:

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Which statement applies to transverse waves?
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Answer:

the waves have a trough

Explanation:

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6 0
3 years ago
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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate
uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: m_{(AcO)_2Cu} = 0.972 g

Volume of the sodium chromate solution: V_{Na_2CrO_4} = 150.0 mL

Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

n_{(AcO)_2Cu} = 0.0053515 mol

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M

8 0
3 years ago
Which of the following acts as a catalyst in catalytic converters?
Serhud [2]

Answer:

B. Metal

Explanation:

The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium. Platinum is used as a reduction catalyst and as an oxidation catalyst. Although platinum is a very active catalyst and widely used, it is very expensive and not suitable for all applications.

Hope it helps plz mark brainlist :)

5 0
3 years ago
Read 2 more answers
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