Answer:
0.0055 mol of N2O5 will remay after 7 min.
Explanation:
The reaction follows a first-order.
Let the concentration of N2O5 after 7 min be y
Rate = Ky = change in concentration of N2O5/time
K is rate constant = 6.82×10^-3 s^-1
Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M
Change in concentration = 0.0117 - y
Time = 7 min = 7×60 = 420 s
6.82×10^-3y = 0.0117 - y/420
0.0117 - y = 420×6.82×10^-3y
0.0117 - y = 2.8644y
0.0117 = 2.8644y + y
0.0117 = 3.8644y
y = 0.0117/3.8644 = 0.00303 M
Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)
Answer: This is true.
Explanation:
It is true because if becomes frozen, then it is physically harder to melt...
I believe they’re both true.
Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
Answer:
How many grams of copper (II) nitrate is formed
