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vivado [14]
4 years ago
15

The decomposition of protein molecules into their basic units will yield: deoxyribonucleic acids ribonucleic acids amino acids n

itric acids
Chemistry
2 answers:
ale4655 [162]4 years ago
5 0

Answer:

amino acids

Explanation:

The building blocks of proteins are amino acids. Two amino acids join with each other with a peptide bond and the release of one molecule of water. This is known as condensation reaction. Amino acids join with each other to form peptides. Peptide chains arrange in 3D conformation to form a protein macromolecule.

liq [111]4 years ago
3 0
Answer would be amino acids.
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5. 1.00 mol HNO3 is treated with 4.47 g of magnesium. Calculate the number of moles of
ruslelena [56]

Answer:

The balanced equation is:

2 HNO3 + Mg ---> Mg(NO3)2 + H2

From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg

Moles of Mg:

Molar mass of Mg = 24 g/mol

Moles = Given mass / Molar Mass

Moles of Mg = 4.47 / 24 = 0.18 moles (approx)

Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed

Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3

Unreacted moles of HNO3 = Total Moles - Moles consumed

Unreacted moles of HNO3 = 0.64 moles (approx)

Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate

From the given options, we can see that 0.632 moles is the closest value to our answer

Therefore, 0.632 moles will remain after the reaction

3 0
4 years ago
Help me please!!!!!!!!?????
kirill [66]

Answer:

i believe its A or D

Explanation:

sorry is its wrong

hope it helps

4 0
3 years ago
Argon-40 undergoes positron emission as shown:
Rudik [331]

Answer:

\frac{40}{18}Ar => \frac{40}{17}Cl + \frac{0}{1}e\\

Explanation:

3 0
4 years ago
How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0
4 years ago
An electron:
Korvikt [17]
It has a negative charge emits energy when it moves to a lower energy orbit from an excited state and it has the same mass as a proton
4 0
4 years ago
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