Answer:
The answer to your question is 160 g of Fe₂O₃
Explanation:
Data
mass of Fe = 112 g
mass of CO = in excess
mass of Fe₂O₃ = ?
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
Process
1.- Calculate the molar mass of Fe₂O₃ and Fe
Molar mass Fe₂O₃ = (56 x 2) + (16 x 3) = 112 + 48 = 160 g
atomic mass of Fe = 56
2.- Use proportions to calculate the mass of Fe₂O₃ needed
160 g of Fe₂O₃ ------------------- 2(56) g of Fe
x g of Fe₂O₃ ------------------ 112 g of Fe
x = (112 x 160) / 2(56)
x = 17920/112
x = 160 g of Fe₂O₃
Answer:
Yes
Explanation:
You add more force behind the cart with the higher mass. Assuming that its higher mass causes it to weigh more.
Answer:
The mass is 1.4701 grams and the moles is 0.01.
Explanation:
Based on the given question, the volume of the solution is 100 ml or 0.1 L and the molarity of the solution is 0.100 M. The moles of the solute (in the given case calcium chloride dihydride (CaCl2. H2O) can be determined by using the formula,
Molarity = moles of solute/volume of solution in liters
Now putting the values we get,
0.100 = moles of solute/0.1000
Moles of solute = 0.100 * 0.1000
= 0.01 moles
The mass of CaCl2.2H2O can be determined by using the formula,
Moles = mass/molar mass
The molar mass of CaCl2.2H2O is 147.01 gram per mole. Now putting the values we get,
0.01 = mass / 147.01
Mass = 147.01 * 0.01
= 1.4701 grams.
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
<h2>Heptene formed is -</h2><h2>

</h2>
Explanation:
The two possibilities when the peroxide is not present
+ HBr →
In presence peroxide,
≡
+ HBr →
- When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
- This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
- One mole of HBr adds to one mole of 1-heptane.
- The structure of heptene formed is -
