Answer:
3.0 moles.
Explanation:
- It is a stichiometry problem.
- The chemical reaction of reacting hydrogen with oxygen to produce water is:
<em>H₂ + 1/2 O₂ → H₂O.</em>
- It is clear that <em><u>1.0 mole of H₂</u></em> reacts with 0.5 mole of O₂ to produce <u><em>1.0 mole of water</em></u>.
- The ratio of the reacting hydrogen to the produced water is 1:1.
∴ The number of moles of water created from reacting 3.0 moles of hydrogen completely with excess oxygen = 3.0 moles.
In order to balance out the Electrons, and give it a net charge of 0, you would need 79 Protons.
Answer:
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B, C, D are compounds, while A and E are just element stand-alones.
Answer:
1.135 M.
Explanation:
- For the reaction: <em>2HI → H₂ + I₂,</em>
The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².
- To solve this problem, we can use the integral law of second-order reactions:
<em>1/[A] = kt + 1/[A₀],</em>
where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),
t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),
[A₀] is the initial concentration of HI ([A₀] = ?? M).
[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).
∵ 1/[A] = kt + 1/[A₀],
∴ 1/[A₀] = 1/[A] - kt
∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.
∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.
<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>
