Answer:
In a nonpolar covalent bond, the atoms share electrons equally with one another.
Explanation:
Nonpolar covalent bonds are a type of bond that occurs when two atoms share a pair of electrons with each other. These shared electrons glue two or more atoms together to form a molecule.
Answer:
![r=0.0341\frac{M}{s}](https://tex.z-dn.net/?f=r%3D0.0341%5Cfrac%7BM%7D%7Bs%7D)
Explanation:
Hello!
In this case, considering the given table, we are able to represent the symbolic rate law as shown below:
![r=k[BF_3]^m[NH_3]^n](https://tex.z-dn.net/?f=r%3Dk%5BBF_3%5D%5Em%5BNH_3%5D%5En)
Thus, by using the following steps, we can find both m (BF3 order of reaction) and n (BF3 order of reaction):
- Experiment 1 and 2 for the calculation of n:
![\frac{r_1}{r_2}=\frac{k[BF_3]_1^m[NH_3]_1^n}{k[BF_3]_2^m[NH_3]_2^n}](https://tex.z-dn.net/?f=%5Cfrac%7Br_1%7D%7Br_2%7D%3D%5Cfrac%7Bk%5BBF_3%5D_1%5Em%5BNH_3%5D_1%5En%7D%7Bk%5BBF_3%5D_2%5Em%5BNH_3%5D_2%5En%7D)
So we plug in to obtain:
![\frac{0.2130}{0.1065}=\frac{k*0.250^m*0.250^n}{k*0.250^m*0.125^n}\\\\2=\frac{0.250^n}{0.125^n}\\\\2=2^n\\\\log(2)=n*log(2)\\\\n=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.2130%7D%7B0.1065%7D%3D%5Cfrac%7Bk%2A0.250%5Em%2A0.250%5En%7D%7Bk%2A0.250%5Em%2A0.125%5En%7D%5C%5C%5C%5C2%3D%5Cfrac%7B0.250%5En%7D%7B0.125%5En%7D%5C%5C%5C%5C2%3D2%5En%5C%5C%5C%5Clog%282%29%3Dn%2Alog%282%29%5C%5C%5C%5Cn%3D1)
So the order of reaction with respect to NH3 is 1.
- Experiment 3 and 4 for the calculation of m
![\frac{r_3}{r_4}=\frac{k[BF_3]_3^m[NH_3]_3^n}{k[BF_3]_4^m[NH_3]_4^n}](https://tex.z-dn.net/?f=%5Cfrac%7Br_3%7D%7Br_4%7D%3D%5Cfrac%7Bk%5BBF_3%5D_3%5Em%5BNH_3%5D_3%5En%7D%7Bk%5BBF_3%5D_4%5Em%5BNH_3%5D_4%5En%7D)
So we plug in to obtain:
![\frac{0.0682}{0.1193}=\frac{k*0.200^m*0.100^n}{k*0.350^m*0.100^n}\\\\0.57=0.57^m\\\\log(0.57)=m*log(0.57)\\\\m=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.0682%7D%7B0.1193%7D%3D%5Cfrac%7Bk%2A0.200%5Em%2A0.100%5En%7D%7Bk%2A0.350%5Em%2A0.100%5En%7D%5C%5C%5C%5C0.57%3D0.57%5Em%5C%5C%5C%5Clog%280.57%29%3Dm%2Alog%280.57%29%5C%5C%5C%5Cm%3D1)
So the order of reaction with respect to BF3 is also 1.
Now, we can compute the rate constant by solving for it on any of the experiments there, say experiment 1:
![k=\frac{r_1}{[BF_3][NH_3]} =\frac{0.2130M/s}{0.250M*0.250M}\\\\k=3.41M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Br_1%7D%7B%5BBF_3%5D%5BNH_3%5D%7D%20%3D%5Cfrac%7B0.2130M%2Fs%7D%7B0.250M%2A0.250M%7D%5C%5C%5C%5Ck%3D3.41M%5E%7B-1%7Ds%5E%7B-1%7D)
Thus, the initial reaction rate for the 0.50M BF3 and 0.020M NH3 is:
![r=3.41M^{-1}s^{-1}*0.50M*0.020M\\\\r=0.0341\frac{M}{s}](https://tex.z-dn.net/?f=r%3D3.41M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.50M%2A0.020M%5C%5C%5C%5Cr%3D0.0341%5Cfrac%7BM%7D%7Bs%7D)
Best regards!
Answer:
T final = 80°C
Explanation:
∴ Q = 18000 cal
∴ m H2O = 300 g
∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K
∴ T1 = 20°C = 293 K
∴ T2 = ?
⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)
⇒ (18000 cal)/(300 cal/K) = T2 - 293 K
⇒ T2 = 293 K + 60 K
⇒ T2 = 353 K (80°C)
Answer: it is just a phrase that mostly everyone uses and they know they ate to much lol
Explanation:
So C12 is the limiting reactant and P4 is the excess
Mass P4 consumed= 0.31 mol X 123.9 g/mol =38.41 g P4 consumed.
Hope i helped