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3 years ago

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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If light with a wavelength of 515 nm is shown on a metal surface, and photoelectrons (electrons ejected from the surface) have a

Debora [2.8K]

The binding energy of the electrons (also known as the work function of the surface) is determined as 2.43 x 10⁻¹⁹ J.

<h3>Binding energy of the electrons</h3>

The binding energy of the electrons is also known as work function of the metal and it is calculated as follows;

Ф = E - K.E

where;

Ф = hf - 86.2 kJ/mol

Ф = hc/λ - 86.2 kJ/mol

Ф = (6.63 x 10⁻³⁴ x 3 x 10⁸ )/515 x 10⁻⁹ - 86.2 kJ/mol

Ф = 3.86 x 10⁻¹⁹ J - (86200 J/mol)/(6.02 x 10²³)

Ф = 3.86 x 10⁻¹⁹ J - 1.43 x 10⁻¹⁹ J

Ф = 2.43 x 10⁻¹⁹ J

Learn more about work function here: brainly.com/question/19427469

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4 0

2 years ago

Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds in the following order: Magne

Drupady [299]

The charge on the monatomic ions of Magnesium, Aluminum, Strontium, and Nitrogen are +2, +3, +2, and -3.

<h3>What is the charge on the ions of atoms of Magnesium, Aluminum, Strontium, Nitrogen?</h3>

The charge on an ion is due to the valency of the atom.

The charge on the ion shows the number of electrons gained or lost by the atom in order to form an ion.

The charge on an ion depends on the group the element belongs to.

Therefore, the charge on the monatomic ions of Magnesium, Aluminum, Strontium, and Nitrogen are +2, +3, +2, and -3.

Learn more about valency at: brainly.com/question/16021551

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8 0

2 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago

How many moles of cabr2 are in 5.0 grams of cabr2?

worty [1.4K]

Moles of CaBr2 = 5/molar mass of CaBr2
= 5/199.886
= 0.025 moles.

Hope this helps!

3 0

3 years ago

Read 2 more answers

Please hurry! And help!!

lyudmila [28]

Answer:

I guess the answer is compound

Explanation:

Coz compound it is on there playing with me again Go ahead and paste it in the class

4 0

3 years ago