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3 years ago

15

Draw a structure for an alcohol that exhibits a molecular ion at M+ = 74 and that produces fragments at m/z = 59, m/z = 56 and m

/z = 45.

1 answer:

pychu [463]3 years ago

6 0

Answer:

Please refer to the attachment below for answer and explanation.

Explanation:

Please refer to the attachment below for answer and explanation.

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What do all four of these types of molecules have in common?

nalin [4]

D. They all contain carbon as an important part of their structure.

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3 years ago

A sample of 0.370 mol of a metal oxide (m2o3) weighs 55.45 g. how many grams of o are in the sample?

Wewaii [24]

0.370 mol metal oxide = 55.45 g

<span>1 mol = 55.45/0.370 = 149.86 g </span>

<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>

<span>there is 48/149.86 * 100% O in the sample </span>

<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>

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How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?

algol [13]

Answer:

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = $\frac{13.0 g}{162.5 g/mol}=0.08 mol$

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

$3\times 0.08 mol=0.24 mol$ of chloride

$Na^++Cl^-\rightarrow NaCl$

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

$\frac{1}{1}\times 0.24 mol=0.24 mol$ of NaCl

1 mole = $N_A=6.022\times 10^{23}$ molecules/ atoms

Number of NaCl molecules in 0.24 moles :

$=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules$

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

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What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?

sp2606 [1]

First step in answering the question is to establish a balanced chemical reaction equation. More specifically, a combustion chemical equation.

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Then using dimension analysis:

$4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) = 0.0184 g O_{2}$

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What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?

Alexxandr [17]

3.3256 Liters

See the image I have shared to you above

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