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3 years ago

The structural level of a protein least affected by a disruption in hydrogen bonding is the

Chemistry

2 answers:

lianna [129]3 years ago

6 0

Answer: Primary level

Explanation:

Marizza181 [45]3 years ago

4 0

Answer:

The structural level of a protein least affected by a disruption in hydrogen bonding is the primary level.

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Which of these elements exhibits chemical behavior similar to that of oxygen? select one:

scoray [572]

Answer:
Sulfur (Option-C) exhibits chemical behavior similar to that of oxygen.

Explanation:
Sulfur has same chemical properties as that of Oxygen because both of them belongs to same group in the periodic table. Also, the similarity of chemical behaviour among the group members is due to same number of electrons in their valence shells.
For examole, the electronic configuration of Oxygen is,

1s², 2s², 2p⁴

There are six valence electrons in the valence shell (i.e. 2) of Oxygen.

Now for Sulfur,
1s², 2s², 2p⁶, 3s², 3p⁴

There are six valence electrons in the valence shell (i.e. 3) of Sulfur.

Therefore, both elements tends to gain 2 electrons in a reaction and form O⁻² and S⁻² respectively.

3 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

Answer: The time taken by the reaction is 84.5 seconds

Explanation:

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago

Write a word equation for the chemical reaction. I’ll brainlist.

Alex

Answer:

potassium hydroxide + sulfuric acid → potassium sulfate + water.

Explanation:

Hope it helps.. if yes, plz mark me as brainliest

5 0

3 years ago

I need help with question 4

meriva

The bat in my opinion...

3 0

3 years ago

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