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Sav [38]
3 years ago
5

What type of reaction breaks the bonds that join the phosphate groups in an ATP molecule?

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

Hydrolysis.

Explanation:

During generation of energy, there is breaking of phosphate bond in the ATP. This breaking of bond is a hydrolysis reaction.

The ATP undergoes hydrolysis, energy is released, and ATP is converted to adenosine diphosphate (ADP) and inorganic phosphate.

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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
At 333 k, which of the pairs of gases below would have the most nearly identical rates of effusion?
rjkz [21]
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
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Hello!
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