Phosphoric acid is a weak acid, while rubidium hydroxide is a strong acid.
H₃PO₄ + RbOH --> Rb₃PO₄ + H₂O
We get Rb₃PO₄ because PO₄ has a charge of 3-, that is PO₄³⁻. Rb has a charge of 1+. You give the subscript of one the charge of the other as this is an ionic compound. So you end up with Rb₃PO₄, a neutral compound.
Now let's balance the equation:
H₃PO₄ + 3RbOH --> Rb₃PO₄ + 3H₂O
Answer:
A. 0.641 L
Explanation:
<em>Given data:</em>
V1 = 2.5 L
T1 = 300.0 K
T2= 77.0 K
<em>To find: </em>
V2= ?
<em>Formula :</em>
By using Charles’ Law
<em>Calculation:</em>
V2 = 2.5 L x 77.0 K / 300.0 K
= 192.5 / 300.0 k
V2 = 0.641 L
If you're asking about the greatest amplitude of visible sunlight, then red light waves.
However if you're talking about the colors just in general, then radio waves.
To solve this problem. use the formula
m1 v1 = m2 v2
where m1 is the molarity of the stock soln
v1 is the volume of the stock soln
m2 is the molarity of the new soln
v2 is the volume of the new soln
m1 v1 = m2 v2
13.5(v2) = (10.5)(1.8)
v2 = (10.5)(1.8) / 13.5
v2 =1.4 L is needed from the 13.5 M HCl stock soln
Answer:
0.5 M
Explanation:
First, let us look at the balanced equation of the reaction.
The solute formed is .
Recall that: mole = molarity x volume
Hence,
50 ml, 1.00 M H2SO4 = 0.05 x 1 = 0.05 mole
50 ml, 2.0 M KOH = 0.05 x 2 = 0.1 mole
From the equation
<em>1 mole of H2SO4 reacts with 2 moles of KOH to give 1 mole of K2SO4.</em>
Hence,
<em>0.05 mole H2SO4 reacting with 0.1 mole KOH will give 0.05 mole </em><em>.</em>
Also recall that: concentration = mole/volume
Total volume of resulting solution = 50 ml + 50 ml = 100 ml or 0.1 liter
Concentration of = mole of /volume of resulting solution
= 0.05/0.1 = 0.5 M
The concentration of the resulting solute = 0.5 M