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Oksana_A [137]
3 years ago
6

The vapor pressure of methanol is 143 mmhg. identify the best reason to explain why methanol spontaneously evaporates in open ai

r at 25.0∘c and standard pressure (760 mmhg). the vapor pressure of methanol is 143 . identify the best reason to explain why methanol spontaneously evaporates in open air at 25.0 and standard pressure (760 ).
Chemistry
1 answer:
goldfiish [28.3K]3 years ago
6 0

Answer/Explanation:

Methanol has a molecular weight (32.04 g/mol), low-boiling point and because of its low boiling point, methanol readily evaporates at room temperature.

Under these specified non-standard conditions, the partial pressure of methanol is lower than its vapor pressure and this explains the reason for the spontaneous evaporation exhibited by methanol.

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What is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest whole-number coeffici
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Option (B) 7

Explanation:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

To know the coefficient of O2 in the above equation, let us balance the equation.

The above equation can be balance as follow:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 3 in front of CO2 as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + H2O(l)

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + 3H2O(l)

There are a total of 4 atoms of O on the left side and a total of 9 atoms on the right side. It can be balance by putting 7/2 in front of O2 as show below:

C3H6O2(l) + 7/2O2(g) → 3CO2(g) + 3H2O(l)

Multiply through by 2

2C3H6O2(l) + 7O2(g) → 6CO2(g) + 6H2O(l)

Now, the equation is balanced.

From the balanced equation above, the coefficient of O2 is 7.

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
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Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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2 years ago
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