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1 year ago

How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?

Chemistry

1 answer:

Kobotan [32]1 year ago

3 0

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

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

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

$0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}$

$\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}$

$\text{Volume of solution (in L)}=2.96L$

Therefore, the volume of solution will be 2.96 liters.

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1 year ago

A penny has a mass of

Ahat [919]

You did not include the questions.

I did some research and found the questions:

<span> What is the mass of 1 mole of pennies? How many moles of pennies have a mass equal to the mass of the moon?

Solutions:

1) mass of 1 mole of pennies

Data: mass of 1 penny = 2.50 g

1 mole = 6.022 * 10^ 23 units

Proportion:

1 penny      6.022 * 10^23 penny
-------------- = ----------------------------
   2.50 g                    x

Solve: x = 6.022 * 10^23 penny * 2.50g / 1 penny = 15.055* 10^23

Since 2.50 has 3 significant figures, the answer must use 3 significant figures => x = 15.1 * 10^ 23 g = 1.51 * 10^24 g

Answer: 1 mol of pennies have a mass of 1.51 * 10^24 g

2) How many moles of pennies have a mass equal to the same mass of the Moon

Convert the mass of the Moon grams: 7.35 * 10^22 kg = 7.35 * 10^ 25 g

       1 mol                            x
---------------------- = ----------------------
1.51 * 10^ 24g          7.35 * 10^ 25 g

=> x = 7.35 * 10^ 25 g * 1 mol / (1.51 * 10^24 g)= 48.7 mol

Answer: 48.7 mol
</span>

5 0

2 years ago

Read 2 more answers

127) Thirty-six colonies grew in nutrient agar from 1.0 ml of undiluted sample in a standard plate count. How many cells were in

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Answer:

Explanation:

Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew colonies then 0.1 ml yielded 6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.

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2 years ago

II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

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Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$

Thus, the mass in excess is:

$Excess=3.86g-0.334g\\\\Excess=3.53g$

Regards!

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2 years ago

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$

So the total mole of chloride ions:

$N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-$

And the total volume by adding the volume of each solution in L:

$V=0.500L+0.425L=0.925L$

Finally, the molarity turns out:

$M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M$

Best regards.

5 0

2 years ago