Answer:
% of n-propyl chloride = 43.48 %
Explanation:
There are 2 secondary hydrogens and 6 primary hydrogens
The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen
probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)
Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)
Total probability = 3.9
% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %
The first step to answering this item is to convert the given temperatures in °F to °C through the equation,
°C = (°F - 32)(5/9)
initial temperature: 72°F
°C = (72 - 32)(5/9) = 22.22°C
final temperature: 145°F
°C = (145 - 32)(5/9) = 62.78°C
Substituting to the equation,
H = mcpdT
H = (43 g)(0.903 J/g°C)(62.78 - 22.22)
H = 1574.82 J
<em>Answer: 1574.82 J</em>
Explanation:
from the equation 1 mole of O2 will give 2 moles of H2O then 6.0 moles of O2 will give x
6.0*2 moles/ 1 mole
= 12 moles
this implies that, 6.0 moles of O2 will give = 12 moles of water
Answer:
We need 92.3 grams of sodium azide
Explanation:
Step 1: Data given
Mass of nitrogen gas = 59.6 grams
Molar mass of nitrogen gas = 28.0 g/mol
Molar mass of sodium azide = 65.0 g/mol
Step 2: The balanced equation
2NaN3 → 2Na + 3N2
Step 3: Calculate moles nitrogen gas
Moles N2 = mass N2 / molar mass N2
Moles N2 = 59.6 grams/ 28.0 g/mol
Moles N2 = 2.13 moles
Step 4: Calculate moles NaN3
for 2 moles NaN3 we'll have 2 moles Na and 3 moles N2
For 2.13 moles N2 we need 2/3* 2.13 = 1.42 moles NaN3
Step 5: Calculate mass NaN3
Mass NaN3 = Moles NaN3 * molar mass NaN3
Mass NaN3 = 1.42 moles * 65.0 g/mol
Mass NaN3 = 92.3 grams
We need 92.3 grams of sodium azide
Answer is: the specific heat capacity of the metal is <span>A) 0.129 J/gK.
</span>m(metal) = 15,1 g.
Q = 48,75 J.
ΔT = 25 K.
Q = C · ΔT · m(metal).
C = Q ÷ ΔT · m(metal).
C = 48,75 J ÷ 25 K · 15,1 g.
C = 0,129 J/g·K.
