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dmitriy555 [2]
3 years ago
12

Which activity removes a large amount of alcohol from the body

Chemistry
2 answers:
lbvjy [14]3 years ago
8 0

Answer:

Running, Peeing.

Explanation:

Georgia [21]3 years ago
6 0

Which activity removes a large amount of alcohol from the body? After alcohol has been consumed and reaches the bloodstream, up to 10% of the alcohol is disposed of through the urine, and 80% to 90% by oxidation in the liver. The average drinker eliminates about 2/3 of a standard drink unit (a shot) per hour.

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What is the formula for potassium citrate and how do you find it?
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Answer:

Potassium citrate (also known as tripotassium citrate) is a potassium salt of citric acid with the molecular formula K3C6H5O7.

Chemical formula: K3C6H5O7

Density: 1.98 g/cm3

Melting point: 180 °C (356 °F; 453 K)

Boiling point: 230 °C (446 °F; 503 K)

Explanation:

7 0
3 years ago
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4HCl(g) + Si (s) ⇌ SiCl4 (l) + 2H2 (g) would be classified as a(n):
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it will be classical as gas

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2H20(g) → 2H2(g) + 02 (8)
fiasKO [112]

Answer:

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5 0
3 years ago
The energy produced when nucleons fuse together is called the: Select the correct answer below:
saveliy_v [14]

Answer:

Option C (nuclear binding energy) is the appropriate choice.

Explanation:

  • At either the nuclear scale, the nuclear binding energy seems to be the energy needed to remove and replace a structure of the atom itself into the characterize elements (to counteract the intense nuclear arsenal).
  • Nuclear warheads (bargaining power) bind everything together neutrons as well as protons within an elementary particle.

Some other options in question aren't relevant to the particular instance. So that the option preceding will also be the right one.

7 0
3 years ago
An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
Anna [14]

Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

5 0
3 years ago
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