2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
Answer:
yes is it yes yes yes yes yes yes yes yes yes yes
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.
Location = nucleus ,
and you can see the total charge on the number that written below the substance, which would be 28+
hope this helps
Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>
