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2 years ago

An unknown substance is found to have a hydrogen concentration of 1.2 * 10 ^ - 2 , what is the pOH? Is it acidic or basic?

Chemistry

1 answer:

Alex_Xolod [135]2 years ago

7 0

Answer:

12.1; acidic

Explanation:

I'm assuming the hydrogen concentration is H+.

So, first we must find pH.

pH = -log[H+]

I simply just plugged in the concentration.

I get 1.9 as the pH. This means that the substance is very acidic since it has a low pH.

Now to find poH.

poH + pH = 14

I simply plugged in once again!

I get 12.1 as the poH. This also determines that the substance is very acidic since it has a high poH.

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Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.

marta [7]

Answer:

Approximately $53.3\; \rm g$ .

Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm H$ : $1.008$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$ :

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$ .

(Rounded to three significant figures.)

6 0

3 years ago

At what time will the pressure of SO₂Cl₂ decline to 0.50 its initial value? Express your answer using two significant figures.

QveST [7]

Answer: The time is 0.69/k seconds

Explanation:

The following integrated first order rate law

ln[SO₂Cl₂] - ln[SO₂Cl₂]₀ = - k×t

where

[SO₂Cl₂] concentration at time t,

[SO₂Cl₂]₀ initial concentration,

k rate constant

Therefore, the time elapsed after a certain concentration variation is given by:

$t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}$

We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:

$p_{(SO_{2}Cl_{2})}V = n_{(SO_{2}Cl_{2})}RT$

$[SO_{2}Cl_{2}]= \frac{n_{(SO_{2}Cl_{2})}}{V}}=\frac{p_{(SO_{2}Cl_{2})}}{RT}}$

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

$t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{p_{(SO_{2}Cl_{2})}} }{k}$

for

$p_{(SO_{2}Cl_{2})}=0.5p_{(SO_{2}Cl_{2})}_{0}$

$t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{0.5p_{(SO_{2}Cl_{2})_{0}}} }{k}$

$t=\frac{ln\frac{1}{0.5} }{k}$

$t=\frac{ln(2)}{k}$

$t=\frac{0.69}{k}}$

7 0

3 years ago

Match the terms with their correct definition:

geniusboy [140]

7 0

3 years ago

The theoretical yield of NaBr from

Lapatulllka [165]

Taking into account definition of percent yield, the percent yield for the reaction is 100%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 FeBr₃ + 3 Na₂S → Fе₂S₃ + 6 NaBr

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

FeBr₃: 2 moles
Na₂S; 3 moles
Fе₂S₃: 1 mole
NaBr: 6 moles

<h3>Moles of NaBr formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 2 moles of FeBr₃ form 6 moles of NaBr, 2.36 moles of FeBr₃ form how many moles of NaBr?

$moles of NaBr=\frac{2.36 moles of FeBr_{3}x6 moles of NaBr }{2 moles of FeBr_{3}}$

moles of NaBr= 7.08 moles

Then, 7.08 moles of NaBr can be produced from 2.36 moles of FeBr₃.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield}x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, being the molar mass of NaBr 102.9 g/mole, you know: