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Nataly [62]
3 years ago
14

What limiting factor is climate considered

Chemistry
1 answer:
gtnhenbr [62]3 years ago
6 0
Climate is considered an abiotic limiting factor because it is non living . hope this helped :))
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Which compound is an isomer of propanoic acid (CH3CH2COOH)?
arlik [135]

Answer:

D) HCOOCH2CH3

Explanation:

An isomer of a compound is defined as a chemical substances with the same formula (That is, same atoms) but in different structures.

For propanoic acid, there are <em>3 atoms of C, 6 atoms of H and 2 atoms of oxygen.</em>

A) CH2CHCOOH . Here you have 3 atoms of C but 4 atoms of H. That means this compound is not an isomer.

B) CH3CH2CH2COOH . Here, there are 4 atoms of C. Thus, is not an isomer.

C) CH3CH(OH)CH2OH. This structure has 3 atoms of C, but 8 atoms of H. Thus, is not an isomer.

D) HCOOCH2CH3. Here, there are 3 atoms of C, 6 atoms of H and 2 atoms of O. Thus, this structure <em>is an isomer of propanoic acid.</em>

5 0
3 years ago
Read 2 more answers
Which describes a chemical reaction? A. Raw eggs are fried in a pan B. Raw eggs are scrambled with a fork C. Raw eggs are placed
Otrada [13]
Raw eggs are scrambled with a fork
8 0
3 years ago
Read 2 more answers
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
3 years ago
A series of chemicals were added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution to produce a precipi
OLEGan [10]

Answer:

First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of HNO_{3}.

Explanation:

In presence of NaCl, AgNO_{3} forms an insoluble precipitate of AgCl.

Reaction: Ag^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)

In presence of NH_{3}, AgCl gets dissolved into solution due to formation of soluble [Ag(NH_{3})_{2}]^{+} complex.

Reaction: AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)

In presence of HNO_{3}, [Ag(NH_{3})_{2}]^{+} complex gets destroyed and free Cl^{-} again reacts with free Ag^{+} to produce insoluble AgCl

Reaction: [Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)

4 0
3 years ago
If you have a heartburn, how you could you safely help the situation
Ket [755]
The answer is:

:Avoid lying down after eating

That should help
7 0
4 years ago
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