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3 years ago

For the closed system below: Cu(s) + 2Ag+(aq) 2Ag(s) + Cu2+(aq) How would you know if the system is at equilibrium?

Chemistry

2 answers:

avanturin [10]3 years ago

7 0

C. the mass of Cu(s) remains constant

satela [25.4K]3 years ago

6 0

its a on plato not c

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Is that the right answer

julsineya [31]

Is what the right answer

8 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Svet_ta [14]

Answer:

$V_1=23.3~mL$

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

$C_1*V_1=C_2*V_2$

Now we can identify the variables:

$C_1=~1.475_M$

$V_1=~?$

$C_2=~0.1374_M$

$V_2=~250.0~mL$

If we plug all the values into the equation:

$1.475_M*V_1=0.1374_M*250.0~mL$

And we solve for $V_1$ :

$V_1=\frac{0.1374_M*250.0~mL}{1.475_M}$

$V_1=23.3~mL$

I hope it helps!

8 0

3 years ago

If the length, width, and height of a box are 8.00 cm, 6.75 cm and 3.50 cm, respectively, what is the volume of the box in units

svp [43]

I think it is 1620 (lxwxh) x 10 to get to millimeters

5 0

4 years ago

Read 2 more answers

ASAP PLEASEE

Lesechka [4]

Answer:

Joe Mama Formula

Explanation:

JoeMamic Acid is the answer

4 0

3 years ago