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3 years ago

Which situation would lead to increased competition

Chemistry

1 answer:

I am Lyosha [343]3 years ago

8 0

I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.

Hope this helps and don't forget to hit that heart :)

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Please help answer this question

Nesterboy [21]

Answer:

hahah animals in the world

4 0

2 years ago

On the addition of HI

victus00 [196]

Answer:

$CH_3-CH(I)-CH(CH_3)-CH_3$

$I-CH_2-CH_2-CH(CH_3)-CH_3$

Explanation:

Given the compound $CH_2=CH-CH(CH_3)-CH_3$

The steps of reaction:

$HI \longrightarrow H^+ + I^-$

$CH_2=CH-CH(CH_3)-CH_3 + H^+ \longrightarrow CH_3-CH^+-CH(CH_3)-CH_3$

But this intermediate product has a resonance:

$CH_3-CH^+-CH(CH_3)-CH_3 \longleftrightarrow C^+H_2-CH2-CH(CH_3)-CH_3$

The reaction with I-

$CH_3-CH^+-CH(CH_3)-CH_3 + I^- \longrightarrow CH_3-CH(I)-CH(CH_3)-CH_3$

$C^+H_2-CH2-CH(CH_3)-CH_3 + I^- \longrightarrow I-CH_2-CH_2-CH(CH_3)-CH_3$

8 0

3 years ago

Organs combine to form … A cells B tissues C organ systems D organisms

vodka [1.7K]

Answer:

The answer is C. Organ systems please brainly me!

Explanation:

3 0

3 years ago

Read 2 more answers

Convert 8.50 moles Ca to atoms

Wewaii [24]

8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

Explanation:

We have to multiply the moles of Ca by the Avogadro's number:

= 6.022×10²³

So the number of atoms:

= 8.5 moles × 6.022×10²³atoms / mol

= 5.1187×10²⁴ atoms

Hence the 8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

4 0

3 years ago

Sucrose, a major product of photosynthesis in green leaves, is synthesized by a battery of enzymes. The substrates for sucrose s

Gwar [14]

Answer:

Sucrose is a disaccharide composed of alpha D gluose and beta D fructose linked together by beta 2,alpha1 glycosidic linkage.

Explanation:

The specificity of glycosidic linkage very much essential to choose the substrate for the synthesis of specific disaccharide.

For example sucrose contain beta 2,alpha1 glycosidic linkage that means the hydroxyl group of anomeric carbon of one monosaccharide(fructose) should remain in beta conformation and the hydroxyl group of other monosaccharide(glucose) should remain in alpha conformation.

5 0

3 years ago