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3 years ago

50.0 grams of KCl is dissolved in water to make a 4.00 L

Chemistry

1 answer:

xeze [42]3 years ago

8 0

Answer:0.1677M

Explanation:

Molarity=moles/volume

Number of moles =mass/molar mass

Once you get the number of moles, you apply it to the molarity formula.

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Which reaction is single replacement?

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3 years ago

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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3 years ago

Which opinion would be an example of an alloy brass or silver or gold or copper

frutty [35]

Answer:

Gold and copper are example of alloy

Explanation:

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2 years ago

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4 years ago

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What is the molarity of 5.60 mol of sodium carbonate in 1500 ml of solution?

FrozenT [24]

Answer:

3.74 M

Explanation:

We know that molarity is moles divided by liters. The first thing to do here is convert your 1500 mL of solution to L. There's 1,000 mL in 1 L, so you need to divide 1500 by 1000:

1500 ÷ 1000 = 1.50

Now you can plug your values into the equation for molarity:

5.60 mol ÷ 1.50 L = 3.74 M

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3 years ago