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IrinaVladis [17]
3 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]3 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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How many MOLES of boron tribromide are present in 3.20 grams of this compound ?
Dmitriy789 [7]

Answer:

1) There are 0.0128 moles of BBr3 present in 3.20 grams of this compound

2) There are 909.35 grams of BBr3 present in 3.63 moles of this compound

Explanation:

<u>Step 1: </u>Given data

Boron tribromide = BBr3

Molar mass of Boron = 10.81 g/mole

Molar mass of Bromide = 79.9 g/mole

Molar mass of Boron tribromide = 10.81 + 3*79.9 = 250.51 g/mole

<u>Step 2:</u> Calculating number of moles

Number of moles = mass / molar mass

Number of moles of BBr3 = 3.20 grams / 250.51 g/mole

Number of moles of BBr3 = 0.0128 moles

<u>Step 3:</u> Calculating mass

If we have 3.63 moles of boron tribromide (= BBr3)

Mass of BBr3 = number of moles of BBr3 * Molar mass of BBr3

Mass of BBr3 = 3.63 moles * 250.51 g/mole

Mass of BBr3 = 909.35 grams

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Which of the following could make a decomposition reaction occur?
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What is the primary type of heat transfer?: A cool breeze blows off the water.
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Answer:

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kari74 [83]
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Explanation : According to the conversion chart, to convert 1° this formula is used;

<span>radians = degrees × π / 180°
</span>
while solving for, One degree  which is equal 0.01745329252 radians:

1° = π/180° = 0.005555556π = 0.01745329252 rad

Answer 2) b = 1' =  = 0.0002907 radians

Explanation : A minute of arc is π/10800 of radian; which is a measurement unit of angular which is equal to 1/60 of 1° degree.

So substituting the value of 1°  as  and on solving we get,

0.01745/60 = 0.0002907 radians.

A minute of arc is <span>π/<span>10800</span></span><span> of a </span>radian.

Answer 3) C= 1'' = 4.8481X 10^{-6}

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