Answer: B) 4.2 M
Explanation:
Molarity = moles of solute/litre of solution
1560mL/1000 = 1.56 litres
6.5/1.56 = x/1
x = 4.2
M of solution = 4.2
<span><span>K_2</span>C<span>O_3</span>(aq)+Ca(N<span>O_3</span><span>)_2</span>(aq)→ ?</span>
If we break these two reactants up into their respective ions, we get...<span><span>
K^+ </span>+ C<span>O^2_3 </span>+ C<span>a^<span>2+ </span></span>+ N<span>O_−3</span></span>
If we combine the anion of one reactant with the cation of the other and vice-versa, we get...<span>
CaC<span>O_3 </span>+ KN<span>O_3</span></span>
Now we need to ask ourselves if either of these is soluble in water. Based on solubility rules, we know that all nitrates are soluble, so the potassium nitrate is. Alternatively, we know that all carbonates are insoluble except those of sodium, potassium, and ammonium; therefore, this calcium carbonate is insoluble.
This is good. It means we have a driving force for the reaction! That driving force is that a precipitate will form. In such a case, a precipitation reaction will occur, and the total equation will be...<span><span>
K_2</span>C<span>O_3</span>(aq) + Ca(N<span>O_3</span><span>)_2</span>(aq) → CaC<span>O_3</span>(s) + 2KN<span>O_3</span>(aq)</span>
To determine the net ionic equation, we need to remove all ions that appear on both sides of the equation in aqueous solution -- these ions are called spectator ions, and do not actually undergo any chemical reaction.
To determine the net ionic equation, let's first rewrite the equation in terms of ions...
2K^+(aq) + CO_3^{2-}(aq) + Ca^{2+}(aq) + 2NO_3^{-}(aq) → Ca^{2+}(s) + CO_3^{2-}(s) + 2K^+(aq) + 2NO_3^-(aq)
The species that appear in aqueous solution on both sides of the equation (spectator ions) are...
<span>
2K^+,NO_3^-</span>
If we remove these spectator ions from the total equation, we will get the net ionic equation...
CO_3^{2-}(aq) + Ca^{2+}(aq) <span>→</span> CaCO_3(s)
Answer:
AS ACCORDING TO THE LAW OF MASS CONSERVATION
REACTANTS =PRODUCTS
THEREFORE,
223.4+96=MASS OF FE2O3
=319.4 FE203
The balanced half-cell equation for the reaction occurring at the anode is H2 ---> 2H(+) + 2e(-)
E<u>xplanation:</u>
- The balanced half-cell equation taking place at the anode is explained below
- The product produced in the reaction in the fuel cell is water.
- H2 ---> 2H(+) + 2e(-)
- In the above reaction, the oxidation state of hydrogen switches from 0 to +1.
- It is becoming oxidized by delivering two electrons at the anode.
- In the fuel cell, hydrogen molecules get oxidized to hydronium ions.Thus half-reaction is the oxidation reaction.
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This is a true statement.
