The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles
<h3>Balanced equation </h3>
2H₂ + O₂ —> 2H₂O
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
<h3>How to determine the mole of oxygen needed </h3>
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
Therefore,
28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen
Thus, 14 moles of oxygen are needed for the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
It is going to be <span>Molar Volume
</span><span>3H2 + N2 --> 2NH3
</span><span> 54.1L*22.4 L/mol H2 , you can find mol of H2, then mol of NH3, and then L of NH3</span>
Answer:
yes
Explanation:
Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.
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 .
E=hc/λ =6.626×10^-34×3 ×10^8 / 3×10^7 × 10^-9 = 6.626×10 ^-24J.
