Answer:
400 mL
Explanation:
Given data:
Mass of barium = 2.17 g
Pressure = 748 mmHg (748/760 = 0.98 atm)
Temperature = 21 °C ( 273+ 21 = 294k)
Milliliters of H₂ evolved = ?
Solution:
chemical equation:
Ba + 2H₂O → Ba(OH)₂ + H₂
Number of moles of barium:
Number of moles = mass/ molar mass
Number of moles = 2.17 g / 137.327 g/mol
Number of moles = 0.016 mol
Now we will compare the moles of barium with H₂.
Ba : H₂
1 : 1
0.016 : 0.016
Milliliters of H₂:
PV = nRT
V = nRT/P
V = 0.016 mol × 0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm
V = 0.39 atm. L/0.98 atm
V = 0.4 L
L to mL
0.4 × 1000 = 400 mL
Answer:
The mass number
Explanation:
The mass number of an atom is its total number of protons and neutrons.
I hope this helps a bit.
I think the answer is -12.7
Answer:
The correct answer is option D.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

Rate of the reaction:
![R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BCO%5D%7D%7Bdt%7D)
Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.
Given rate expression of the reaction:
![R = k[NO2]^2[CO]](https://tex.z-dn.net/?f=R%20%3D%20k%5BNO2%5D%5E2%5BCO%5D)
Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'
![R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R](https://tex.z-dn.net/?f=R%27%3Dk%282%5Ctimes%20%5BNO_2%5D%29%5E2%282%5Ctimes%20%5BCO%5D%29%3D8%5Ctimes%20k%5BNO2%5D%5E2%5BCO%5D%3D8R)
Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.
Hence, none of the given statements are true.