A nanometer (nm) is 1/billion of a meter (1×10^-9 m). Write this number:

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask

tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M$

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

$Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L$

0.823 Molar is the molarity of the solution.

6 0

3 years ago

A 2.87 g sample of carbon reacts with hydrogen to form 3.41 g of car fuel. What is the empirical formula for the car fuel?

olga_2 [115]

CH₇ is the empirical formula of the car fuel.

Explanation:

To find the empirical formula we use the following algorithm.

First divide each mass the the molar weight of each element:

for carbon 2.87 / 12 = 0.239

for hydrogen 3.41 / 2 = 1.705

And now divide each quantity by the lowest number which is 0.239:

for carbon 0.239 / 0.239 = 1

for hydrogen 1.705 / 0.239 = 7.13 ≈ 7

The empirical formula of the car fuel is CH₇.

I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.

Learn more about:

empirical formula

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5 0

3 years ago

Can a solution with undissolved solute be supersaturated

Helen [10]

Answer: A supersaturated solution will not contain undissolved solute because the undissolved solute will be indicative of saturated solution.

Explanation:

A supersaturated solution is the one that consists of more than the maximum concentration of the solute in the solvent that is being dissolved at a given temperature. A saturated solution is the one in which the maximum concentration of solute has been dissolved in the solvent and no additional solute can be dissolved further.

According to the given statement, a solution with undissolved solute is a saturated solution rather a supersaturated solution.

5 0

3 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0

3 years ago

Which type of relationship is responsible for teaching you values and how to communicate?

Blizzard [7]

The answer would be family.

8 0

3 years ago