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3 years ago

Anyone know this I Need help

Chemistry

1 answer:

aleksandr82 [10.1K]3 years ago

5 0

Answer:

which one?

Explanation:

do you need help on all?

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A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

Stells [14]

Answer:

e. T₂= 4T₁

Explanation:

Initially, we have a number of moles (n₁) a gas sample at a certain pressure (P), temperature (T₁) and volume (V). We can relate these variables through the ideal gas equation.

P . V = n₁ . R . T₁

where,

R is the ideal gas constant

We can rearrange this equation like:

$T_{1}=\frac{P.V}{n_{1}.R}$

If only one fourth of the initial molecules remain n₂ = 1/4 n₁. The new temperature (T₂) assuming pressure and temperature remain constant is:

$P.V=n_{2}.R.T_{2}=\frac{1}{4} n_{1}.R.T_{2}\\\frac{P.V}{n_{1}.R} =\frac{1}{4} T_{2}\\T_{1}=\frac{1}{4} T_{2}\\T_{2}=4.T_{1}$

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3 years ago

Students were asked to identify the location of elements in the

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Answer:амьсгал боогдов гэж найдаж байна

Explanation:

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3 years ago

9. Which of the following is true?

wlad13 [49]

The answer should be C

6 0

3 years ago

What is the difference between boiling point and normal boiling point?

Softa [21]

Answer:

The boiling point is the temperature at which the vapor pressure equals the pressure of gas.

The normal boiling point is the temperature at which the vapor pressure equals one atmosphere

Explanation:

5 0

3 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago