The molarity of Barium Hydroxide is 0.289 M.
<u>Explanation:</u>
We have to write the balanced equation as,
Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O
We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,
V1M1 = 2 V2M2
Here V1 and M1 are the volume and molarity of nitric acid
V2 and M2 are the volume and molarity of Barium hydroxide.
So the molarity of Ba(OH)₂, can be found as,

= 0.289 M
Answer:
Oxide of M is
and sulfate of 
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:


Moles of hydrogen gas produced = 0.01225 mol

Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

x = 2.9 ≈ 3


Formulas for the oxide and sulfate of M will be:
Oxide of M is
and sulfate of
.
The percentage of yield was 777.78%
<u>Explanation:</u>
We have the equation,
Be
[s] + 2
HCl
[aq] → BeCl
2(aq] +
H
2(g] ↑ Be
(s] +
2
HCl
[aq] → BeCl
2(aq] +
H
2(g]
↑
To find the percent yield we have the formula
Percentage of Yield= what you actually get/ what you should theoretically get x 100
=3.5 g/0.45 g 100
= 777.78 %
The percentage of yield was 777.78%
Answer:
Hope this helps
Explanation:
Potential energy diagrams represent the energy transfer in chemical reactions in a diagram called a potential energy graph and/or a reaction progress curve. A potential energy diagram shows the adjustment in potential energy of a system as reactants are changed.
Redox Reaction is an ionic bond