Genetic variation is important because it allows natural selection to increase or decrease frequency of alleles already in the population
Answer:
A. N2S3
yep u r right
Explanation:
Dinitrogen Trisulfide N2S3 Molecular Weight -- EndMemo.
Answer:
676mmHg
Explanation:
Using the formula;
P1/T1 = P2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 725.0mmHg
P2 = ?
T1 = 20°C = 20 + 273 = 293K
T2 = 0°C = 0 + 273 = 273K
Using P1/T1 = P2/T2
725/293 = P2/273
Cross multiply
725 × 273 = 293 × P2
197925 = 293P2
P2 = 197925 ÷ 293
P2 = 676mmHg.
The resulting pressure is 676mmHg
Answer:
The value of dissociation constant of the monoprotic acid is 
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)
Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)
![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)
The value of dissociation constant of the monoprotic acid is .
Answer:
Explanation:
Your strategy here will be to
use the chemical formula of carbon dioxide to find the number of molecules of
CO
2
that would contain that many atoms of oxygen
use Avogadro's constant to convert the number of molecules to moles of carbon dioxide
use the molar mass of carbon dioxide to convert the moles to grams
So, you know that one molecule of carbon dioxide contains
one atom of carbon,
1
×
C
two atoms of oxygen,
2
×
O
This means that the given number of atoms of oxygen would correspond to
4.8
⋅
10
22
atoms O
⋅
1 molecule CO
2
2
atoms O
=
2.4
⋅
10
22
molecules CO
2
Now, one mole of any molecular substance contains exactly
6.022
⋅
10
22
molecules of that substance -- this is known as Avogadro's constant.
In your case, the sample of carbon dioxide molecules contains
2.4
⋅
10
22
molecules CO
2
⋅
1 mole CO
2
6.022
⋅
10
23
molecules CO
2
=
0.03985 moles CO
2
Finally, carbon dioxide has a molar mass of
44.01 g mol
−
1
, which means that your sample will have a mass of
0.03985
moles CO
2
⋅
44.01 g
1
mole CO
2
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
a
a
1.8 g
a
a
∣
∣
−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the number of atoms of oxygen present in the sample.
