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3 years ago

The solubility of a gas is .55 g/l at 8.0 atm pressure. what will be the solubility of the gas at 5.0 atm partial pressure?

1 answer:

8 0

Henry's law states that:

P1/C1 = P2/C2

Where P = Pressure, and C = Solubility
In this case, P1 = 8.0 atm, C1 = 0.55 g/l, P2 = 5.0 atm, C2 =??

Therefore,
C2 = (P2*C1)/P1 = (5*0.55)/8 = 0.344 g/l

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3 years ago

Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm

Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining $\rho_A:$ Ethyl alcohol density and $\rho_G:$ Glycerin density , we can write:

$\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3$

Simplifying:

$\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)$

On the other hand:

$h_1 + h_2 = \Delta h + h_3$

Rearranging:

$h_1 - \Delta h = h_3 - h_2 (2)$

Replacing (2) in (1):

$\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)$

Rearranging:

$\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h$

Data:

$h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}$

$\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}} = \Delta h$

$7.5 cm = \Delta h$

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3 years ago

How much work is done in holding a 20 N sack of potatoes while waiting in line at the

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Answer: A

Explanation:

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2 years ago

You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the

podryga [215]

The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
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kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4

3 0

3 years ago

Read 2 more answers

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$