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3 years ago

What is the [H+] in a solution with pOH of 0.253?

Chemistry

2 answers:

grandymaker [24]3 years ago

7 0

Answer: pOH of the solution is 13.41.

Explanation:-

pH is the measure of acidity or alkalinity. It is also defined as negative logarithm of hydrogen ion concentration.

$pH=-log[H^+]$

$pOH=-log[OH^-]$

$pH+pOH=14$

Given: $[H^+]$ = 0.253

Thus $pH=-log[0.253]$

$pH=0.59$

$0.59+pOH=14$

$pOH=14-0.59=13.41$

sergey [27]3 years ago

4 0

PH + pOH = 14

pH + 0.253 = 14

pH = 14 - 0.253

pH = 13.747

[ H+] = 10 ^ -pH

[ H+ ] = 10 ^- 13.747

[ H+ ] = 1.790x10⁻¹⁴ M

hope this helps!

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An aqueous solution contains 32.7% kcl (weight/weight %). how many grams of water are contained in 100 g of this solution

igor_vitrenko [27]

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Please help ill medal and fan!

krok68 [10]

Solutions for your questions are the following:
1. remaining amount is equal to:960 g : 100% = 30 g : xx = 30 g * 100% / 960 g
= 3.125%
= 0.03125
Now, we use this formula to calculate the number of half-lives:(1/2)ⁿ = x,
so,(1/2)ⁿ = 0.03125
to calculate n, use this equation:
n*log(1/2) = log(0.03125) n = log(0.03125)/log(1/2)
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Then
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3 years ago

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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