Answer: b.) they tend to lose electrons to gain stability
Explanation:
Answer:
84.8 mL
Explanation:
From the question given above, the following data were obtained:
Mass of CuNO₃ = 3.53 g
Molarity of CuNO₃ = 0.330 M
Volume of solution =?
Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:
Mass of CuNO₃ = 3.53 g
Molar mass of CuNO₃ = 63.5 + 14 + (16×3)
= 63.5 + 14 + 48
= 125.5 g/mol
Mole of CuNO₃ =?
Mole = mass / Molar mass
Mole of CuNO₃ = 3.53 / 125.5
Mole of CuNO₃ = 0.028 moles
Next, we shall determine the volume of the solution. This can be obtained as follow:
Molarity of CuNO₃ = 0.330 M
Mole of CuNO₃ = 0.028 moles
Volume of solution =?
Molarity = mole /Volume
0.330 = 0.028 / Volume
Cross multiply
0.330 × Volume = 0.028
Divide both side by 0.330
Volume = 0.028 / 0.330
Volume = 0.0848 L
Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.0848 L = 0.0848 L × 1000 mL / 1 L
0.0848 L = 84.8 mL
Therefore, the volume of the solution is 84.8 mL.
<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)
