ANSWER : 108 + 10 = 118
118 + ( 5 + 3 )
118 + 8 = <u>1</u><u>2</u><u>6</u>
<u>=</u><u> </u><u>1</u><u>2</u><u>6</u><u> </u><u>g</u>
Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
0.43 atm
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 0.95 atm
Initial volume (V₁) = 0.55 L
Final volume (V₂) = 1.22 L
Final pressure (P₂) =?
The final pressure of the gas can be obtained by using the Boyle's law equation as follow:
P₁V₁ = P₂V₂
0.95 × 0.55 = P₂ × 1.22
0.5225 = P₂ × 1.22
Divide both side by 1.22
P₂ = 0.5225 / 1.22
P₂ = 0.43 atm
Therefore, the final pressure of the gas is 0.43 atm
Not all DNA codes for proteins
Explanation:
- DNA (Deoxy Ribonucleic acid) is the genetic material present in all the eukaryotes.
- An experiment conducted in the 1960s proved that most parts of DNA are composed of long strings of repeated sequence which does not code for proteins.
- The process involves in coding for a protein are the transcription of DNA to RNA, then translation followed by protein synthesis. There are enzymes involved in these processes.
- As per the latest study, it states that only 1 per cent of genes code for proteins rest 99 per cent is called the non-coding DNA.
Answer:
2.4 x 1027 g
Explanation:
the mass of 1 mole of brick is 2.4 x 1027 g.
