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Ulleksa [173]
2 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]2 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]2 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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Read 2 more answers
A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect
juin [17]

Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

b. The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

Kinetic energy = K = 4.25×10⁻⁵ J

a. What work was done by the electric force?

This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

V2 = -3.03 * 10³ V

The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

4 0
3 years ago
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