<u>We are given:</u>
M1 = 3 Molar V1 = 80 mL
M2 = x Molar V2 = 100 mL
<u>Finding the molarity:</u>
We know that:
M₁V₁ = M₂V₂
where V can be in any units
(3)(80) = (x)(100)
x = 240/100 [dividing both sides by 100]
x = 2.4 Molar
Answer: The gas generated by two antacid tablets has a smaller volume.
Explanation:
Since the antiacid is the limiting reagent, we know that the more tablets there are, the more gas there will be.
This means that there will be more gas generated by the four antiacid tablets when compared to the two antiacid tablets, which gives us that the gas generated by the two antiacid tablets has a smaller volume.
Answer:
The answer to your question is V2 = 4.97 l
Explanation:
Data
Volume 1 = V1 = 4.40 L Volume 2 =
Temperature 1 = T1 = 19°C Temperature 2 = T2 = 37°C
Pressure 1 = P1 = 783 mmHg Pressure 2 = 735 mmHg
Process
1.- Convert temperature to °K
T1 = 19 + 273 = 292°K
T2 = 37 + 273 = 310°K
2.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (783 x 4.40 x 310) / (292 x 735)
-Simplification
V2 = 1068012 / 214620
-Result
V2 = 4.97 l
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
