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Gennadij [26K]
3 years ago
8

Oxidation number of the sulfur and chloride in the sulfur tetrachloride?​

Chemistry
1 answer:
yuradex [85]3 years ago
6 0

Answer:

SCl4

oxidation state of sulphur=+4

oxidation state of chlorine=-4

oxidation state of one chlorine atom=-1

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What are the major difference between potential and kinetic energy?​
Sholpan [36]

Answer:

The energy of a body or a system with respect to the motion of the body or of the particles in the system. Potential energy is the stored energy in an object or system because of its position or configuration. Kinetic energy of an object is relative to other moving and stationary objects in its immediate environment.

6 0
3 years ago
Which of the following is not a primary function of the products of the pentose phosphate pathway? a. To maintain levels of NADP
zlopas [31]

Answer: The correct answer is D.

Explanation:

As you can see in every step of the pathway, ATP is not generated directly. Although, in the oxidative phase, molecules of NADPH are produced and these can be used for synthesize ATP indirectly; the NADPH molecules are hihgly demanded in tissues that carry out constantly synthesis of fatty acids, such as liver.

Therefore the correct answer is D.

6 0
3 years ago
Formic acid (HCO2H) has a dissociation constant of 1.8 _ 104 M. The acid dissociates 1:1. What is the [H+] if 0.1 mole of formic
Umnica [9.8K]
<span>Use the equation for Ka => 1.8 x 10 ^ (-4) = [H+][HCO2-}, these 2 concentrations are equal so, </span>
<span>1.8 x 10 ^ (-4) = [H+]^2 </span>

<span>and [H+] = 4.2 × 10^-3 M </span>
7 0
4 years ago
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Levart [38]

Answer:organic

Explanation:

7 0
3 years ago
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A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027
irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

        x              _________  25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

7 0
4 years ago
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