Molar mass
HNO₃ = 63.02 g/mol
N₂O = 44.013 g/mol
The balanced equation is :
<span>4 Zn + 10 HNO₃ = 4 Zn(NO₃)₂ + N₂O + 5 H₂O
</span>
10 * 63.02 g HNO₃-----------> 44.013 g N₂O
? HNO₃ -----------------> 8.75 g N₂O
Mass HNO₃ = 8.75 * 10 * 63.02 / 44.013
Mass HNO₃ = 125 g
hope this helps!
Answer:
is the value of the rate constant.
Explanation:

Let the order of the reaction be x.
The rate law of the reaction can be written as:
![R=k[H_2O_2]^x](https://tex.z-dn.net/?f=R%3Dk%5BH_2O_2%5D%5Ex)
1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds to 60 seconds.

..[1]
2. Rate of the reaction when concentration changes from 0.697 M to 0.566 M in 240 seconds to 360 seconds.

..[2]
[1] ÷ [2]
![\frac{0.00308 M/s}{0.00227 M/s}=\frac{k[0.697 M]^x}{k[0.236M]^x}](https://tex.z-dn.net/?f=%5Cfrac%7B0.00308%20M%2Fs%7D%7B0.00227%20M%2Fs%7D%3D%5Cfrac%7Bk%5B0.697%20M%5D%5Ex%7D%7Bk%5B0.236M%5D%5Ex%7D)
Solving fro x:
x = 0.92 ≈ 1
![R=k[H_2O_2]^1](https://tex.z-dn.net/?f=R%3Dk%5BH_2O_2%5D%5E1)
![0.00308 M/s=k[0.697 M]^1](https://tex.z-dn.net/?f=0.00308%20M%2Fs%3Dk%5B0.697%20M%5D%5E1)
![k=\frac{0.00308 M/s}{[0.697 M]^1}=0.00442 s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.00308%20M%2Fs%7D%7B%5B0.697%20M%5D%5E1%7D%3D0.00442%20s%5E%7B-1%7D)
is the value of the rate constant.
Answer: to see if the matter is a compound, mixture, or element.
Explanation:
can you please help with my most recent question :)
Answer: A. or C. (most likely C.)
Explanation:
All molecules of a compound have the same properties. ... All molecules of a compound have one type of atom.
Jeez, they r both true.....go with A. or C.
Hope this helps!
Answer:
n = 2 moles (1 sig-fig)
Explanation:
Using the Ideal Gas Law equation (PV = nRT), solve for n (= moles) and substitute data for ...
pressure = P(atm) = 100atm
volume =V(liters) = 50L
gas constant = R = 0.08206L·atm/mol·K
temperature = T(Kelvin) = °C + 273 = (35 + 273)K = 308K
PV = nRT => n = PV/RT = (100atm)(50L)/(0.08206L·atm/mol·K)(308K)
∴ n(moles) = 1.978moles ≅ 2 moles gas (1 sig-fig) per volume data (= 50L) that has only 1 sig-fig. (Rule => for multiplication & division computations round final answer to the measured data having the least number of sig-figs).