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Burka [1]
2 years ago
11

Countercurrent Flow

Chemistry
1 answer:
Nikitich [7]2 years ago
7 0

Answer: All options (I, II, III, IV, V)

Explanation: The answer is all options because they never reach equilibrium or intersect at a region as the both decrease at the same rate.

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A solution is prepared by dissolving 27.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
bagirrra123 [75]
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is 
     % by mass = (mass of the solute / mass of the solution)*100 
                        = mass of solute / (mass of the solute + mass of the solvent)*100
                        = (27.7 g CaCl2 / 27.7g + 375g) * 100 
                        = 6.88
5 0
3 years ago
Read 2 more answers
Explain why chloro acetic acid<br>strong and than acetic acid<br>​
Basile [38]

Chloroacetic acid is stronger than acetic acid because of the electron-withdrawing effect of chlorine. This effect is caused by the electronegativity.

7 0
3 years ago
How are parentheses used in chemical formulas
Liula [17]
They make the symbols individual
6 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
Using the chemical equation below, how many moles of Al2O3 produce when 105 grams of Al are reacted? 2Al+Fe2O3 —&gt; Al2O3+2Fe
topjm [15]

Answer:

1.94

Explanation:

moles = mass/Mr so you have 3.889 moles of aluminium, but because the ratio in the equation is 2:1 you need to halve it.

4 0
2 years ago
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