Answer:
Initial rate of the reaction when concentration of hydrogen gas is doubled will be
.
Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Initial rate of the reaction = R = 
![R = k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%20%3D%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
![4.0\times 10^5 M/s=k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E5%20M%2Fs%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
The initial rate of the reaction when concentration of hydrogen gas is doubled : R'
![[H_2]'=2[H_2]](https://tex.z-dn.net/?f=%5BH_2%5D%27%3D2%5BH_2%5D)
![R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3](https://tex.z-dn.net/?f=R%27%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%27%5E3%3Dk%5Ctimes%20%5BN_2%5D%5B2H_2%5D%5E3)
![R'=8\times k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%27%3D8%5Ctimes%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)

Initial rate of the reaction when concentration of hydrogen gas is doubled will be
.
Answer:
Explanation:
<u>1) Data:</u>
a) V = 93.90 ml
b) T = 28°C
c) P₁ = 744 mmHg
d) P₂ = 28.25 mmHg
d) n = ?
<u>2) Conversion of units</u>
a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter
b) T = 28°C = 28 + 273.15 K = 301.15 K
c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm
d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm
<u>3) Chemical principles and formulae</u>
a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.
b) Ideal gas equation: pV = nRT
<u>4) Solution:</u>
a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm
b) Moles of hygrogen gas:
pV = nRT ⇒ n = pV / (RT) =
n = (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =
n = 0.00358 mol (which is rounded to 3 significant figures) ← answer
Answer:
hard and brittle
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