Explanation:
Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s)
1 0.100 2.7 × 10−7
2 0.080 2.2 × 10−7
3 0.055 1.5 × 10−7
4 0.044 1.2 × 10−7
a) Choose the rate law for the decomposition of CS2.
Comparing equations 1 and 3, reducing the initial concentration by almost half (from 0.100 to 0.055) leads too the rate of reaction to be reduced by almost half (from 2.7 × 10−7 to 1.5 × 10−7).
This signifies that the reaction is a first order reaction.
Rate = k [CS2]
(b) Calculate the average value of the rate constant.
Taking equation 1.
Rate = k [CS2]
k = Rate / [CS2]
k = 0.100 / (2.7 × 10−7) = 0.037 x 10^8 = 3.7 x 10^6s-1
Answer:
The answer is C: has at least three oxidation states.
Explanation:
you're welcome
How much of a product is made depends on the limiting reagent. there are different ways to determine the limiting reagent and this depends on the professor way of teaching it. I will use the must common one. so keep in mind that the way to solve this problem may vary.
first we convert each grams to moles using the molar mass of the molecules. then convert moles to moles of asked molecule (H2O) using the mole-mole ration, and moles of water to grams using the molar mass of water.
molar mass of Na= 23.0 g/mol
molar mass of H₂O= 18.0 g/mol
ratios (based on the balanced equation coefficients)
2 mol Na= 1 mol H₂
2 mol H₂O= 1 mol H₂
calculations:
![120. g Na \frac{1 mol Na}{23.0 g} x \frac{1 mol H_{2} }{2 mol Na} x \frac{2.02 g}{1 mol H_2} = 5.27 g H_2](https://tex.z-dn.net/?f=120.%20g%20Na%20%5Cfrac%7B1%20mol%20Na%7D%7B23.0%20g%7D%20x%20%5Cfrac%7B1%20mol%20H_%7B2%7D%20%7D%7B2%20mol%20Na%7D%20x%20%5Cfrac%7B2.02%20g%7D%7B1%20mol%20H_2%7D%20%3D%205.27%20g%20H_2)
![80. g H_2 O \frac{1 mol H_2 O}{18.0 g} x \frac{1 mo H_2}{2 mol H_2 O} x \frac{2.02 g}{1 mol H_2} = 4.49 g H_2](https://tex.z-dn.net/?f=80.%20g%20H_2%20O%20%5Cfrac%7B1%20mol%20H_2%20O%7D%7B18.0%20g%7D%20x%20%5Cfrac%7B1%20mo%20H_2%7D%7B2%20mol%20H_2%20O%7D%20x%20%5Cfrac%7B2.02%20g%7D%7B1%20mol%20H_2%7D%20%3D%204.49%20g%20H_2)
since H₂O gives less of the product than the other one. H₂O is the limiting reagent and the answer is
4.49 g
Answer:
I'm not really sure if you're interested in the electron dot diagram of the potassium and bromine atoms, or of potassium bromide,
KBr, so I'll show you both. You can use this example to find the electron dot diagram of hydrogen bromide, HBr.
Explanation: