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2 years ago

What is the percent, by mass, of water in MgSO4.2H20

Chemistry

1 answer:

iris [78.8K]2 years ago

8 0

Answer:51.1%

Explanation:

Mass percent : It is defined as the mass of the given component present in the total mass of the compound. Formula used : First we have to calculate the mass of and . Mass of = 18 g/mole Mass of = 7 × 18 g/mole = 126 g/mole Mass of = 246.47 g/mole Now put all the given values in the above formula, we get the mass percent of in . Therefore, the mass percent of in is, 51.1%

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Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1

tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0

3 years ago

Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio

Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

50.48 % O = 50.48 grams

49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0

3 years ago

Identify the type of error in the following scenario. SCENARIO: A group is trying to find the volume of a given liquid. To do th

7nadin3 [17]

Answer:

Error of parallax it usually occurs when the cylinder is above or below the eye level,and thus resulting in differences In reading the millimeters

5 0

3 years ago

Which of the following is an example of a chemical change? A. melting solid gold B. burning hydrogen gas C. breaking a sheet of

cestrela7 [59]

The answer is B. Burning hydrogen gas

5 0

3 years ago

230 in scientific notation

rewona [7]

230 in scientific notation is:

2.3 x 10^2

7 0

3 years ago