Initially, weigh the correct amount of 
calculated from the formula.
Number of moles =
As 1 mole of is given and molar mass of is 
, then
1 mole of = 
Mass of in g = 
Thus,
1 mole of = 58.44 g.
Now, the weighed amount of i.e. 58.44 g is added to a 1 liter container and then add small amount of water in the container to dissolve the salt. After that, fill the container with distilled water to the graduation mark or until the total volume reaches 1 liter.
Thus, option (c) is correct i.e. first put one mole of salt in the container then add the water with stirring till the total volume reaches 1 liter.
Answer:
Mg(s) + 2H⁺ → Mg²⁺(aq) + H₂.
Explanation:
- To obtain the overall reaction, we sum the two half-reactions and omit the similar species in the reactants and products sides.
oxidation reaction:
Mg(s) → Mg²⁺(aq) + 2e.
Reduction reaction:
2H⁺ + 2e → H₂.
- So, we add the two half-reactios and obtain the overall reaction:
<em>Mg(s) + 2H⁺ → Mg²⁺(aq) + H₂.</em>
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Answer:
That's simple it's the caca of 54 32 x (5) b ÷ 5683jvl done so your answer is 0
5.00111 moles in 45.05g of br
Answer:
0.0585 M
Explanation:
- Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)
First we <u>calculate the inital number of moles of each reagent</u>, using the <em>given volumes and concentrations</em>:
- 0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂
- 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl
Then we <u>calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl</u>, using the <em>stoichiometric coefficients of the reaction</em>:
- 16.0 mmol NaCl *
= 8.00 mmol Pb(NO₃)₂
Now we <u>calculate the remaining number of Pb(NO₃)₂ moles after the reaction</u>:
- 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂
Finally we <em>divide the number of moles by the final volume</em> to <u>calculate the concentration</u>:
- 5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M
