Basically, this problem asks you to convert units of volume from cm³ to μL. Cubic centimeters is the cubic volume equivalent of milliliters or mL. According to the metric system for L, microliter is 10⁻⁶ of a liter, while milliliter is 10⁻³ of a liter. Therefore, the solution is written as
(1.12×10⁻²³ cm³)*(1 mL/1 cm³)*(10⁻³ L/ 1 mL)*(1μL/10⁻⁶ L) = 1.12×10⁻²⁰ μL
Answer:
![-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}](https://tex.z-dn.net/?f=-r_%7BA%7D%3Dk%5Ctimes%5BBF_3%5D%5E%7B1%7D%5Ctimes%5BNH_3%5D%5E%7B1%7D)
Explanation:
The rate law of a chemical reaction is given by
![-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}](https://tex.z-dn.net/?f=-r_%7BA%7D%3Dk%5Ctimes%5BBF_3%5D%5E%7B%5Calpha%7D%5Ctimes%5BNH_3%5D%5E%7B%5Cbeta%7D)
This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found
Between experiments 1 and 2
![\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta](https://tex.z-dn.net/?f=%5Cfrac%7B-r_%7BA1%7D%7D%7B%7B-r%7D_%7BA2%7D%7D%3D%5Cleft%28%5Cfrac%7B%5Cleft%5BNH_3%5Cright%5D_1%7D%7B%5Cleft%5BNH_3%5Cright%5D_2%7D%5Cright%29%5E%5Cbeta)
Then the expression for the calculation of 
![\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cfrac%7Bln%5Cfrac%7B-r_%7BA1%7D%7D%7B-r_%7BA2%7D%7D%7D%7Bln%5Cleft%28%5Cfrac%7B%5Cleft%5BNH_3%5Cright%5D_1%7D%7B%5Cleft%5BNH_3%5Cright%5D_2%7D%5Cright%29%7D%3D%5Cfrac%7Bln%5Cfrac%7B0.2130%7D%7B0.1065%7D%7D%7Bln%5Cleft%28%5Cfrac%7B0.250%7D%7B0.125%7D%5Cright%29%7D)
Resolving
Doing the same between experiments 3 and 4 the expression for 
is
![\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7Bln%5Cfrac%7B-r_%7BA3%7D%7D%7B-r_%7BA4%7D%7D%7D%7Bln%5Cleft%28%5Cfrac%7B%5Cleft%5BBF_3%5Cright%5D_3%7D%7B%5Cleft%5BBF_3%5Cright%5D_4%7D%5Cright%29%7D%3D%5Cfrac%7Bln%5Cfrac%7B0.0682%7D%7B0.1193%7D%7D%7Bln%5Cleft%28%5Cfrac%7B0.200%7D%7B0.350%7D%5Cright%29%7D)
Resolving
This means that the rate law for this reaction is
I'm not positive, but I believe it would be one near the Rocky Mountains.
Answer:
Explanation:
To convert from grams to atoms, first divide by the molar mass, the multiply by 6.022*10^23.
To convert from moles to mass, multiply by the molar mass of the element.
Hope this helps!
-Emma Victoria
Answer:
0.038 g of reactant
Explanation:
Data given:
Heat release for each gram of reactant consumption = 36.2 kJ/g
mass of reactant that release 1360 J of heat = ?
Solution:
As 36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J
As we know
1 KJ = 1000 J
So
36.2 kJ = 36.2 x 1000 = 36200 J
So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release
Apply unity formula
36200 J of heat release ≅ 1 gram of reactant
1360 J of heat release ≅ X gram of reactant
Do cross multiplication
X gram of reactant = 1 g x 1360 J / 36200 J
X gram of reactant = 0.038 g
So 0.038 g of reactant will produce 1360 J of heat.
