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3 years ago

Enter a balanced chemical equation for the reaction between potassium metal and liquid water.

Chemistry

1 answer:

Oksanka [162]3 years ago

6 0

Answer:

2K (s) + 2H₂O (l) → 2KOH (aq) + H₂(g)↑

Explanation:

1) Word equation (given)

Potassium + water → ?

2) Kind of reaction

Potassium is an alkali metal.

This is a single replacement reaction.

Alkali metals is the group of metals in the group 1 of the periodic table. These are the most active metals, meaning that they react violently with water releasing hydrogen gas.

Alkali metals can lose an electron to form compounds with oxidation number +1.

The general equation for the reaction of an alkali metal and water is:

Metal + Water → hydroxide + gaseous hydrogen

X + H₂O → XOH(aq) + H₂ (g) ↑

3) Chemical equation

K (s) + H₂O (l) → KOH(aq) + H₂(g)↑

4) Balance

2K (s) + 2H₂O (l) → 2KOH(aq) + H₂(g)↑

There is the same number of each atom on both sides: 2 K in the left and in the right, 4 H in the left and in the right, and 2 oxygen in the left and in the right. So, the equation is balanced.

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4.8 moles of Kl are dissolved in 850 mL of water. What is the molarity of the solutions?

AURORKA [14]

Hey there!

Molarity = 4.8/850 x 1000 =5.647mol/lit or 5.65M

Hope this helps you dear :)
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Answer:

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3 years ago

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Answer:

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3 years ago

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Neutralizing an olympic size swimming pool is conceptually very similar to performaing a massive titration experiment. Suppose a

MrRissso [65]

Answer:

6,97x10⁻³ gallons

Explanation:

pH is defined as:

pH = -log [H⁺]

Thus, you need to have, in the end:

10⁻⁷ = [H⁺]

And you have, in the first:

$10^{-9,33}$ = [H⁺]

The volume of swimming pool is:

700'000 galllons × $\frac{3,78541 L}{1 gallon}$ = 2649787 L

Thus, the moles of H⁺ in the first and in the end are:

First:

$10^{-9,33}mol/L$ × 2649787L = 1,24x10⁻³ moles

End:

$10^{-7}mol/L$ × 2649787L = 0,265 moles

Thus, the moles of H⁺ you need to add are:

0,265 - 1,24x10⁻³ = 0,26376 moles

These moles comes from 10M HCl, thus, the volume in gallons you need to add are:

$0,26376moles*\frac{1L}{10moles}* \frac{1gallon}{3,78541L}$ =

6,97x10⁻³ gallons

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3 years ago

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

So, $N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}$

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

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3 years ago