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Rainbow [258]
3 years ago
5

- If I initially have a gas at a pressure of 379 mmHg, a volume of 467 mL, and a temperature of

Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
5 0

Answer:

301.89 mL

Explanation:

- Use combined gas law formula

- Rearrange that formula to fit values then plug them in.

- Be sure to change C to K

Hope that helped! Let me know if you need further explanation.

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Perform the following
Sever21 [200]

Answer:

8.97998

Explanation:

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3 years ago
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Anti-acids because they contain Alkaline ion that chemically neutralizes stomach acid
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Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
C,H. + 020 H2O + CO2<br> How many grams of O, are needed to<br> combust 55.0 grams of CHỞ?
Studentka2010 [4]

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3 years ago
You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that
Alecsey [184]

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

                                     = 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

                        = 0.005 g

B = \frac{0.005}{0.05}

  = 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

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