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Daniel [21]
2 years ago
5

How many grams are in 3 moles of Oxygen?

Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
4 0

Answer:

48 g

Explanation:

Molar mass of oxygen is 16 g/mol

3 mol x 16g/mol

48 g

You might be interested in
Convert 67 inches into a feet
fiasKO [112]

Answer:

Explanation:every feet has 12in so divde, 67/12  so its 5ft 7/12

4 0
3 years ago
Calculate the pH of a buffer solution prepared by mixing 60.0 mL of 1.00 M lactic acid and 25.0 mL of 1.00 M sodium lactate.
marshall27 [118]
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:

pH = pKa + log[(salt/acid]

Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid). 

Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L

Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M

Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
4 0
3 years ago
you have been observing an insect that defends itself from enemies by secreting a caustic liquid. analysis of the liquid shows i
Alexus [3.1K]

pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions

[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate

[HCOOH] = 1.45 M - 0.015 M

[HCOOH] = 1.435 M; equilibrium concentration of formic acid

pKa = -logKa

pKa = -log 1.8×10⁻⁴ M

pKa = 3.74

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 3.74 + log (0.015 M/1.435 M)

pH = 3.74 - 1.98

pH = 1.76

More about buffer: brainly.com/question/4177791

#SPJ4

4 0
2 years ago
What is the mass (grams) of salt in 10.0 m' of ocean water? ball park-4x10's (1.000 molsalt -58.44 g salt, 1.0 L ocean water -0.
koban [17]

Answer:

3.5 × 10⁵ g of salt

Explanation:

<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>

We have this data:

  • 1.000 mol salt is equal to 58.44 g salt
  • 1.0 L of ocean water contains 0.60 mol of salt

We will need the following relations:

  • 1 L = 1dm³
  • 1 m³ = 10³ dm³

We can use proportions:

10.0m^{3} .\frac{10^{3}dm^{3}  }{1m^{3} } .\frac{1L}{1dm^{3} } .\frac{0.60molSalt}{1.0L} .\frac{58.44gSalt}{1molSalt} =3.5 \times 10^{5} gSalt

8 0
3 years ago
The accepted value for the mass of a moonrock is 46.37 g. In an experiment the rock is measured to be 47.25 g. What is the perce
Artist 52 [7]

Answer:

The answer is

<h2>2.00 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual measurement = 46.37 g

error = 47.25 - 46.37 = 0.88

The percentage error of the measurement is

P(\%) =  \frac{0.88}{46.37}  \times 100 \\  = 1.89777873...

We have the final answer as

<h3>2.00 %</h3>

Hope this helps you

4 0
3 years ago
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