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Vikki [24]
3 years ago
5

The concentration of dye in Solution A is 26.609 M. You have 13 mL of water at your disposal to make the dilutions. The solution

is diluted twice, to make Solutions B and C. In the first dilution, 5 parts of Solution A is diluted with 12 parts water to make Solution B. In the second dilution, 7 parts of Solution B is diluted with 4 parts water to make Solution C. What is the concentration of dye in Solution C? Provide your response to three digits after the decimal.
Chemistry
1 answer:
exis [7]3 years ago
8 0

Answer:

6,613 M

Explanation:

Dilution is the process of reducing the concentration of a solute in solution, mixing initial solution with more solvent.

The concentration of Solution B is:

23,881 M ×  = 9,552 M

Because the initial eight parts are diluted to 12+8 parts.

Thus, concentration of solution C is:

9,552 M ×  = 6,613 M

I hope it helps!

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strojnjashka [21]
-log (1×10^-12) is how you calculate the pOH which in this case is 12
7 0
3 years ago
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What mass of carbon dioxide is produced from the complete combustion of 5.30x10-3 g of methane?
pshichka [43]

Answer:

1.45 x 10⁻² g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

5.30 x 10⁻³ g CH₄          1 mole              1 mole CO₂          44.007 g
---------------------------  x  ----------------  x  ---------------------  x  -----------------  =
                                      16.043 g            1 mole CH₄            1 mole

=  0.0145 g CO₂

=  1.45 x 10⁻² g CO₂

8 0
2 years ago
______ have no definite shape, but definite volume
NemiM [27]

Answer:

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Explaination:

3 0
3 years ago
Equal moles of H2, N2, O2, and He are placed into separate containers at the same temperature. Assuming each gas behaves ideally
lbvjy [14]

Answer:

They would all exhibit the same pressure.

Explanation:

Since the same number of mole of each gas is placed in different containers, it means the gas will occupy the same volume.

Now, the gases were observed at the same temperature. This means they will all have the same pressure as their volume is the same.

Now we can further understand this by doing a simple calculation as follow:

Assumptions:

For H2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, H2 has a pressure of 1 atm.

For N2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, N2 has a pressure of 1 atm

For O2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, O2 has a pressure of 1 atm

For He:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, He has a pressure of 1 atm.

From the above illustrations we can see that the gases have the same pressure since they have the same number of mole, volume and were observed at the same temperature.

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