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ololo11 [35]
4 years ago
14

The first student recognizes this as the first step of the acetoacetic ester synthesis. he treats the starting material with sod

ium methoxide followed by methyl iodide. he isolates compound a, but 1h nmr analysis shows this is not the desired material. elemental analysis shows it has the same molecular formula as the 2-methyl-3-oxobutanoic acid. what is compound a?
Chemistry
1 answer:
krek1111 [17]4 years ago
8 0
The first student treats the starting material which is the 3-oxobutanoic acid with sodium methoxide which is followed by methyl iodide and form compound A. Analysis 1HNMR it shows that this is not the desired material.
Compound A has same 2-methyl-3- oxobutanoate acid as molecular formula.
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The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3
Oksanka [162]

Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

n=3.84mol

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

3 0
3 years ago
What is the molar mass for CaWO4
djverab [1.8K]

Answer:

Calcium tungstate

287.93 g/mol

8 0
2 years ago
0.378g grams of a compound is made up of 0.273g of Mg and 0.105g nitrogen. Find its empirical formula.
Harrizon [31]

Answer:

The empirical formula of the compound is Mg_3N_2.

Explanation:

Mass of magnesium in compound = 0.273 g

Mass of nitrogen in the compound = 0.105 g

Moles of magnesium =\frac{0.273 g}{24.305 g/mol}=0.0112 mol

Moles of nitrogen = \frac{0.105 g}{14.0067 g/mol}=0.00750 mol

Form empirical formula divides the lowest value of moles of an element present with all the moles of elements present.

Magnesium =\frac{0.0112 mol}{0.00750 mol}=1.5

Nitrogen =\frac{0.00750mol}{0.00750 mol}=1

The empirical formula of a compound:

Mg_{1.5}N_1=Mg_{\frac{15}{10}}N_1=Mg_{15}N_{10}=Mg_3N_2

The empirical formula of the compound is Mg_3N_2.

3 0
3 years ago
What evidence indicates that nitric acid (HNO3) is a strong acid?​
GREYUIT [131]

Answer:

The presence of hydrogen ions

Explanation:

Nitric acid is formed when hydrogen ions combine with nitrogen oxide ions leading to formation of a strong acid. Wheareas for strong bases the hydroxyl iins must be present, for strong acid, the hydrogen ions contribute to the acidity. Therefore, from the formula of nitric acid, it has hydrogen ions, an indicator of a strong acid.

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3 years ago
Which statement is correct about the theory of combustion? (1 point)
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Answer:

i need the answer

Explanation:

6 0
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