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Gwar [14]
2 years ago
8

Will give brainliest if correct. Which of these bonds should be classified as a polar covalent bond?

Chemistry
1 answer:
quester [9]2 years ago
4 0

Answer:

The answer is O-H.

Explanation:

This is because when you subtract their EN values you get 1.4, and that is in the range of the polar covalent bond values.

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<span>Valence electrons are those in the outer most orbital shell (the valence shell). Elements in the same column (group) have the same number of valence electrons. Hydrogen is in Group 1. If you were to look at a periodic table of the elements, Lithium is right under Hydrogen. You are correct, the answer is D. Lithium.</span>
6 0
3 years ago
Help ill give brainliest need help with number 8
kotegsom [21]

Answer:

the sun goes away and takes heat with it

Explanation:

5 0
3 years ago
Read 2 more answers
Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g
satela [25.4K]

Answer:

5.6gNa_2SO_4

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4

Best regards.

5 0
3 years ago
14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km
emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0
3 years ago
What products are always made when any carbon-based fuel
Elodia [21]
Answer:
d) carbon dioxide and water
8 0
2 years ago
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