NaBr (aq) + _H3PO4 (aq) → _Na3PO4(aq) + _HBr (aq)<br>
3,1,1,3<br>
1,2,3,1<br>
3,1,3,1<br>
1,1,3,3
BaLLatris [955]
Answer:
A 3,1,1,3
Explanation:
hope the picture help u to understand :)
Answer:
1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
Explanation:
Let's consider the dissolving equations for the following compounds.
1. Rubidium hydroxide
RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Sodium carbonate
Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. Ammonium selenite
(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
<span>Acetone has a density of 784 kg/mÂł (this you can look up).
This equals 784.000 g/mÂł (kilo means 1000).
Which can be converted to 0.784 g/cm3 (1 m3 is 1.000.000 cm3).
A cm3 is exactly 1 ml. So 0.784 g/cm3 = 0.784 g/ml.
We have 6.68 gram of acetone. 6.68 g / 0.784 g/ml = 8.52 ml.
So the volume of 6.68 grams of acetone is 8.52 ml.</span>
Answer:
Explanation:
At the cathode
In case of molten AgI
Silver will be collected
In case of molten LiI
lithium will be collected
in case of aqueous LiI,
hydrogen gas will be collected as reduction potential of H⁺ is more than Li⁺
in case of aqueous AgI,
Silver will be obtained at cathode because reduction potential of silver is more than H⁺
At the Anode
In case of molten NaBr
Bromine will be collected
In case of molten NaF
Fluorine will be collected
in case of aqueous NaBr ,
Bromine will be collected as reduction potential of Br⁻ is less than O⁻²
in case of aqueous NaF ,
oxygen will be obtained because reduction potential of F⁻ is more than O⁻² .
