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4vir4ik [10]
3 years ago
8

What are some questions about light pollution?

Chemistry
1 answer:
devlian [24]3 years ago
6 0

Answer:

is there a way of reducing light pollution at the edges of a forest work?

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The answer is c, relying on renewable energy sources
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3 years ago
This chart lists four elements from the periodic table.
aksik [14]
Iodine and Calcium is the correct answer
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3 years ago
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Urea makes up a significant amount of urine, and is the chemical that the body uses to rid cells of unwanted carbon dioxide and
Ira Lisetskai [31]
Urea is highly soluble in water. When it is allowed to dissolve in water in the presence of heat, it will yield ammonia and carbon dioxide. The reaction is shown below:

<span>NH2-CO-NH2 + H2O </span>⇒ 2 NH3 + CO2

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3 0
3 years ago
[OH-] for a solution is
solong [7]

Answer:

B = basic

Explanation:

Given data:

[OH⁻] = 5.35×10⁻⁴M

pH = ?

Solution:

pOH = -log[OH⁻]

pOH = - [5.35×10⁻⁴]

pOH = 3.272

it is known that,

pH + pOH = 14

pH = 14- pOH

pH = 14 - 3.272

pH = 10.728

The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.

8 0
3 years ago
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Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq) (acidic solution) express your answer as a chemical equation. identify all of the phases in
Rashid [163]
<span>Answer: Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
4 0
3 years ago
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