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IrinaVladis [17]
2 years ago
6

A chemist has 2.0 mol of methanol (CH3OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?

16 30 32 64
Chemistry
1 answer:
rodikova [14]2 years ago
3 0

Answer:

\boxed {\boxed {\sf D. \ 64 \ grams }}

Explanation:

Given the moles, we are asked to find the mass of a sample.

We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.

\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}

Multiply by the given number of moles, which is 2.0

2.0 \ mol \ CH_3OH *\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}

The moles of methanol will cancel each other out.

2.0 \ *\frac{32 \ g \ CH_3OH}{1 }

The denominator of 1 can be ignored.

2.0 * 32 \ g\ CH_3OH

Multiply.

64 \ g \ CH_3OH

There are 64 grams of methanol in the sample.

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In this question, the <span>patient needs to be given exactly 500 ml of a 5.0%. The content of the glucose should be:
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2. Mass of AuCl₃ in 750 mL

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8 0
2 years ago
You wish to make a buffer with pH 7.0. You combine 0.060 grams of acetic acid and 14.59 grams of sodium acetate and add water to
aleksandr82 [10.1K]

Answer:

The pH of the buffer is 7.0 and this pH is not useful to pH 7.0

Explanation:

The pH of a buffer is obtained by using H-H equation:

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<em>Where pH is the pH of the buffer</em>

<em>The pKa of acetic acid is 4.74.</em>

<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>

<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>

<em />

Replacing:

pH = 4.74 + log [0.1779mol] / [0.001mol]

<em>pH = 6.99 ≈ 7.0</em>

<em />

The pH of the buffer is 7.0

But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,

this pH is not useful to pH 7.0

<em />

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2 years ago
Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions ar
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We have to know the number of Na⁺ ions in the unit cell.

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3 0
2 years ago
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