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2 years ago

How much 0.02 m kmno4 solution should be needed if the solutions tested have a composition of 3% h2o2 by mass?

1 answer:

6 0

When the concentration is expressed in mass percentage, that means there is 3 g of solvent H₂O₂ in 100 grams of the solution. Then, that means the remaining amount of solute is 97 g. We use the value of molarity (moles/liters) to determine the amount of solution in liters, denoted as V. The solution is as follows:

0.02 mol KMnO4/L solution * 158.034 g KMnO4/mol * V = 97 g KMnO4

Solving for V,
V = 30.69 L

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A group or family represents the *blank* on the periodic table

alekssr [168]

The answer is vertical column

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2 years ago

An unknown compound contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 2.77 g sample of the compound

alisha [4.7K]

The molecular formula of the unknown compound is determined as C₉H₁₁NO₂.

<h3>Molecular formula of the compound</h3>

The molecular formula is calculated as follows;

CHNO + O₂ ------------> CO₂ + H₂O

Mass of carbon, C: = (6.64 x 12)/44 = 1.81 g in 2.77 g sample

Mass of hydrogen, H: = (1.67 x 2)/18 = 0.186 g in 2.77 g sample

Mass of Nitrogen, N: = (2.77 x 0.143)/1.69 = 0.234 g

Mass of oxygen, O: = 2.77 g - 1.81 g - 0.186 g - 0.234 g = 0.54 g

<h3>molar ratio of the elements: </h3>

C = 1.81 g = 0.15 mol

H = 0.186 g = 0.186 mol

N = 0.234 g = 0.017 mol

O = 0.54 g = 0.0337 mol

divide through with the smallest number of moles (0.017 mol);

C = 9

H = 11

N = 1

O = 2

Molecular formula = C₉H₁₁NO₂

Check the molar mass of the compound = (9 x 12) + (11 x 1) + (14) + (2 x 16) = 165 g/mol

Thus, the molecular formula of the unknown compound is determined as C₉H₁₁NO₂.

Learn more about molecular mass here: brainly.com/question/21334167

#SPJ1

4 0

1 year ago

What is the overall ionic equation for zinc + silver nitrate?

3241004551 [841]

Answer:

Zn(NO₃)₂

Explanation:

this single replacement reaction will produce silver metal, Ag , and aqueous zinc nitrate, Zn(NO3)2 . Zinc is above silver is the metal reactivity series, so it will replace silver in silver nitrate

7 0

2 years ago

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]

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

<u>Explanation:</u>

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

4 0

2 years ago

What happens to an atom if it loses an electron?

jolli1 [7]

Answer:

When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons. In the opposite case when an atom gains electron(s) it becomes negatively charged (more electrons than protons).

Explanation:

7 0

2 years ago