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cupoosta [38]
3 years ago
12

How many water molecules are formed during the formation of a triglyceride?

Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
4 0
It consists of three individual fatty acids bound together in a single Large molecular
You might be interested in
In making a pizza, which process involves a chemical
Natasha_Volkova [10]

Answer:

baking the dough to form the crust

8 0
3 years ago
When na and s undergo a combination reaction, what is the chemical formula of the product?
kvasek [131]
The Chemica formula of the product is Na2S which is called sodium sulfide. remember that sodium is a metal and all compounds containing a metal are named with the stock system. 
5 0
3 years ago
Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh
Ann [662]

Answer:

See explanation and image attached

Explanation:

The  Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi.  R is an alkyl or aryl group.

They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.

In this particular instance, we intend to synthesize propylcyclohexane. The structure of the  lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.

7 0
3 years ago
How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?
timama [110]

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

3 0
3 years ago
Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o
Gnom [1K]

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

7 0
3 years ago
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