Answer:
69.036 g of Compound is required to obtain 48.45 g of Oxygen.
Explanation:
Let us assume that the total mass of the compound is 100 g. So it means that this compound weighing 100 is composed 70.18 % of Oxygen or among 100 g of this compound 70.18 g is constituted by oxygen only. Hence, we can make a relation as that,
70.18 g O is present in = 100 g of a Compound
So,
48.45 g of O will be present in = X g of a Compound
Solving for X,
X = 100 g × 48.45 g ÷ 70.18 g
X = 69.036 g of Compound
Answer:
For every temperature you use the formula given by the question.
at 3°C you have:
Reaction rate = (1000/0.2)/138.5 = 36 mg/L/sec
at 24°C you have:
Reaction rate = (1000/0.2)/34.2 = 146 mg/L/sec
at 40°C you have:
Reaction rate = (1000/0.2)/26.3 = 190 mg/L/sec
at 65°C you have:
Reaction rate = (1000/0.2)/14.2 = 352 mg/L/sec
Explanation:
Answer:
see below
Explanation:
3°C: 36
24°C: 146
40°C: 190
65°C: 352
Full table: 145
Eight pieces: 173
Tiny pieces: 216
3
8
16
36
Answer:
36
Explanation:
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Answer:
36
Explanation:
This is a simple answer, but I assure you that it is 100% correct.
Hope this helps you.
Answer:
-232 °C
Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data:
m = 20.0 g
p = 6.0 atm
V = 0.40 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations:
1. Moles of N₂
2. Temperature of N₂
3. Convert the temperature to Celsius
T = (41.0 - 273.15) °C = -232 °C