The struggle for existence against other members of an organisms species is

The Great Plains are also referred to as____.

Gnesinka [82]

Answer:

The interior plains

5 0

2 years ago

Read 2 more answers

To make 2.00 L of 0.300 M sulfuric acid, how many mL of a 1.00 M stock solution should be used?

Tamiku [17]

We are given with
M1 = 1.00 M

M2 = 0.300 M
V2 = 2.00 L

We are asked to get V1

Using material balance
M1 V1 = M2 V2
Substituting the given values
1.00 V1 = 0.300 M (2.00 L)
V1 = 0.600 L or 600 mL
THe volume needed is 600 mL

3 0

2 years ago

a solution is prepared by dissolving 55g of CaCl2 into 300g of water. if the Kf of water is -1.86 c/m, shat is the freezing poin

Zepler [3.9K]

Answer:

Relation between , molality and temperature is as follows.

T =

It is also known as depression between freezing point where, i is the Van't Hoff factor.

Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.

i for = 3

i for glucose = 1

i for NaCl = 2

Depression in freezing point will have a negative sign. Therefore, d

depression in freezing point for the given species is as follows.

=

Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.

Glucose < NaCl <

Explanation:

3 0

2 years ago

Argon atom can exist freely in nature, why

lana [24]

Answer:

Argon can exist freely in nature because it has a full octet of electrons the way its found in the nature is the same way its found in periodic table of element in vast amouts of stabilization.

7 0

2 years ago

When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0

2 years ago