All categories

3 years ago

The rainforest contains what percentage of all of the species in the world?

Chemistry

2 answers:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

I think 50%

Explanation:

Yuliya22 [10]3 years ago

7 0

I believe it’s about 50% :)

You might be interested in

Bruh i need help and someone just used the last one for points plz help

ankoles [38]

Answer: 5000

Explanation:

6 0

3 years ago

Calculate the mass of 3 moles of clorine

Vitek1552 [10]

Moles Chlorine to grams = 35.453 grams.
moles Chlorine to grams = 70.906 grams.
moles Chlorine to grams = 106.359 grams.
moles Chlorine to grams = 141.812 grams.
moles Chlorine to grams = 177.265 grams.
moles Chlorine to grams = 212.718 grams.

4 0

3 years ago

What comes in contact with Sodium Hydroxide to form soap and glycerine it is a chemical change because you cannot reverse this c

svetoff [14.1K]

Answer:

Your answer is triglycerides. Hope this helps.

8 0

3 years ago

Which member of each pair is more soluble in diethyl ether? Why?<br> (c) MgBr₂(s) or CH₃CH₂MgBr(s)

White raven [17]

CH3CH2MgBr is more soluble in diethyl ether .

We know that polar solvent dissolve in polar solvent very perfectly . as diethyl ether is a polar solvent so it have dipole -dipole interaction .

Hence the compound with similar interaction can dissolve in diethyl ether .

Here , MgBr2 is an ionic compound . there is ion-ion interactions occurs which is not similar to dipole -dipole interaction in diethyl ether .hence the solubility of MgBr2 in diethyl ether is less .

but in case of CH3CH2MgBr there are both polar and nonpolar end .CH3CH2 is the nonpolar end and MgBr is the polar end .

thus with the nonpolar end solute interact using depression forces and with polar end solute interact using dipole-dipole interaction . so CH3CH2MgBr is more soluble .

Learn more about polar solvent here :

brainly.com/question/3184550

#SPJ4

4 0

2 years ago

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

3 years ago