Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
Best regards!
Answer:
1.1
Explanation:
The slope of a line can be calculated using the values of the x and y corrdinates. The equation is given as;
M = ΔY / ΔX = Y2 - Y1 / X2 - X1
From the points;
(25.6cm³ , 28.16g) - (X1, Y1)
(17.3cm³, 19.03g) - (X2, Y2)
Inserting the values into the equation;
M = 19.03 - 28.16 / 17.3 - 25.6
M = -9.13 / -8.3 = 1.1
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
![\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
=21 torr/min
The balanced chemical reaction is,
The rate of disappearance of 
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of 
= ![-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
The rate of formation of 
= ![\frac{1}{2}\frac{d[NOCl]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D)
As we know that,
=21 torr/min
So,
![-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2021torr%2Fmin%3D10.5torr%2Fmin)
And,
![\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D)
![\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%3D21torr%2Fmin)
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.
Answer:
0.01504 mol/dm3
Explanation:
Calcium Hydroxide - Ca(OH)2
Hydrochloric acid - HCl
The reaction between both is given as;
2HCl + Ca(OH)2 --> CaCl2 + 2H2O
2 : 1
The relationship between the concentration and volume is given as;
CaVa / CbVb = na / nb
where a = acid (HCl) and b = base Ca(OH)2
Inserting the values;
0.04 * 18.8 / (Cb * 25) = 2 / 1
0.752 / 25Cb = 2
Cb = 0.752 / 50 = 0.01504
I know for a fact that a good conductor transfer electricity well,.. so my guess would be that the answer is true.
