<u>Answer:</u> The volume of water required is 398 mL
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
We are given:
Mass of solute (manganese (II) nitrate tetrahydrate) = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Molarity of solution = 0.16 M
Putting values in above equation, we get:
Hence, the volume of water required is 398 mL
Speed is how fast something is going, velocity is speed and distance, and acceleration is how fast something is speeding up.
The 7 parts of the water cycle would be precipitation runoff evaporation condensation than precipitation again and runoff again 3>
Answer:
Pb (NO₃)₂(aq) + 2KCl(aq) → 2KNO₃(aq) + PbCl₂(s)
Explanation:
In given chemical equation the aqueous lead (II) nitrate react with aqueous potassium chloride and form aqueous potassium nitrate and lead chloride.
Chemical equation:
Pb (NO₃)₂(aq) + KCl(aq) → KNO₃(aq) + PbCl₂(s)
Balanced chemical equation:
Pb (NO₃)₂(aq) + 2KCl(aq) → 2KNO₃(aq) + PbCl₂(s)
ionic equation:
Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺(aq) + 2Cl⁻ (aq) → 2NO₃⁻(aq) + 2K⁺(aq) + PbCl₂(s)
Net ionic equation:
Pb²⁺ (aq) + 2Cl⁻ (aq) → PbCl₂(s)
The NO₃⁻(aq) and K⁺(aq) are spectator ions that's why these are not written in net ionic equation. The PbCl₂ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be __6__.
Explanation:
From the redox equation, we can see that NO₃⁻ is reduced to NO (from oxidation state +5 to +2), whereas I⁻ is oxidized to I₂ (from oxidation state -1 to 0). The half reactions are balanced with H⁺ (acidic solution), as follows:
Reduction : 2 x (NO₃⁻(aq) + 3 e- + 4 H⁺ → NO(g) + 2 H₂O)
Oxidation : 3 x (2 I⁻(aq) → I₂(s) + 2 e-)
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Total equation: 6 I⁻(aq) + 2 NO₃⁻(aq)+ 8 H⁺ → 3 I₂(s) + 2 NO(g) + 4 H₂O
That is the redox equation with the smallest whole number coefficients.
Accordin to this, the coefficient for the iodide ion (I⁻) is: 6.
