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3 years ago

Calculate the molarity of 0.700 mol of na2s in 1.05 l of solution.

1 answer:

6 0

Molarity is the amount of solute molecule (in moles) per 1L of solvent. In this case, the solute is 0.7mol Na2S and the solvent volume is 1.05L. Since the unit in this problem is already mol and L then you don't need to do any conversion of the units. The calculation would be:

molarity = mol of solute / (1L/ volume of solvent)
molarity = 0.7 mol/ (1L/ 1.05L)= 0.67M

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Gnesinka [82]

Answer:

proactively develop ROE as well as responding to requests for additional ROE measures

Explanation:

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3 years ago

What is the formula for copper(1) chlorate?

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Answer:

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3 years ago

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

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3 years ago

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alexandr1967 [171]

Answer:

Option B and A

Explanation:

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1 year ago

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Answer:

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3 years ago