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3 years ago

Caculate the mass of a solid with a density of 14.2 g/cm cubed and a volume of 350 cm cubed

Chemistry

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

The mass of solid with density 14.2 g/cm3 and volume 350 cm3 is 4.97 Kg.

Explanation:

Formula for density is given below :

$density = \frac{mass}{volume}$

$volume = 350 cm^{3}$

$density = 14.2 g/cm^{3}$

Insert the values in the formula

$14.2 = \frac{mass}{350}$

on cross multiplication :

$mass = density\times volume$

$mass = 14.2\times 350$

$mass = 4970 g$

1 g = 0.001 Kg

$mass =4970\times 0.001 Kg$

mass = 4.97 Kg

Note : Check the units of volume and density carefully

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I have no clue how to even start this, any help is appreciated!

monitta

Answer:

Option D. Carbon monoxide, 4.56 Moles

Explanation:

The balanced equation for the reaction is given below:

Fe₂O₃ + 3CO —> 2Fe + 3CO₂

From the balanced equation above,

1 mole of Fe₂O₃ reacted with 3 moles of CO to produce 3 moles of CO₂.

1. Determination of the limiting reactant.

From the balanced equation above,

1 mole of Fe₂O₃ reacted with 3 moles of CO.

Therefore, 2.54 moles of Fe₂O₃ will react with = 2.54 × 3 = 7.62 moles of CO.

From the calculation made above, we can see that it will take higher amount of CO (i.e 7.62 moles) than what was given (i.e 4.56 moles) to react completely with 2.54 moles of Fe₂O₃. Therefore, CO is the limiting reactant and Fe₂O₃ is the excess reactant.

2. Determination of the maximum amount of carbon dioxide CO₂ produced.

In this case, the limiting reactant will be used since all of it is consumed by reaction.

The limiting reactant is CO and the maximum amount of CO₂ produced can be obtained as follow:

From the balanced equation above,

3 moles of CO reacted to produce 3 moles of CO₂.

Therefore, 4.56 moles of CO will also react to produce 4.56 moles of CO₂.

Thus, the maximum amount of CO₂ produced is 4.56 moles

Summary:

Limiting reactant => carbon monoxide, CO

Maximum amount of carbon dioxide, CO₂ produced = 4.56 moles

Thus, option D gives the correct answer to the question.

5 0

3 years ago

A 1.1000 gram carbonate sample chosen from Li2CO3, K2CO3, Na2CO3 and CaCO3 was reacted with H2SO4 and was found to lose 0.3497 g

Stels [109]

Answer:

For 1: The percentage composition of $CO_2$ in unknown carbonate sample is 31.79 %

For 2: The unknown carbonate sample is potassium carbonate.

Explanation:

To calculate the percentage composition of carbon dioxide in sample, we use the equation:

$\%\text{ composition of }CO_2=\frac{\text{Mass of }CO_2}{\text{Mass of sample}}\times 100$ .....(1)

For 1:

Mass of sample = 1.1000 g

Mass of carbon dioxide = 0.3497 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{0.3497g}{1.1000g}\times 100=31.79\%$

Hence, the percentage composition of $CO_2$ in unknown carbonate sample is 31.79 %

For 2:

We are given:

Mass of carbon dioxide = $[1\times 12.01)+(2\times 16.00)]=44.01g$

For lithium carbonate:

Mass of lithium carbonate= 73.892 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{73.892g}\times 100=59.55\%$

The percent composition of carbon dioxide in lithium carbonate is 59.55 %

For potassium carbonate:

Mass of potassium carbonate= 138.21 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{138.21g}\times 100=31.84\%$

The percent composition of carbon dioxide in potassium carbonate is 31.84 %

For sodium carbonate:

Mass of sodium carbonate= 105.99 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{105.99g}\times 100=41.52\%$

The percent composition of carbon dioxide in sodium carbonate is 41.52 %

For calcium carbonate:

Mass of calcium carbonate = 100.09 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:

$\%\text{ composition of }CO_2=\frac{44.01g}{100.09g}\times 100=43.97\%$

The percent composition of carbon dioxide in calcium carbonate is 43.97 %

As, the percent composition of carbon dioxide in the unknown sample matches the percent composition of carbon dioxide in potassium carbonate.

Hence, the unknown carbonate sample is potassium carbonate.

7 0

3 years ago

13.How many grams of phosphorus (P4) are needed to completely consume 79.2 L of chlorine gas according to the following reaction

BabaBlast [244]

Answer:

73.2g

Explanation:

The reaction expression is given as:

P₄ + 6Cl₂ → 4PCl₃

Given parameters:

Volume of chlorine gas = 79.2L

Unknown:

Mass of Phosphorus needed = ?

Solution:

To solve this problem, let us find the number of moles of the chlorine gas.

Since the condition of the reaction is at STP;

22.4L of gas is contained in 1 mole

79.2L of chlorine gas will contain $\frac{79.2}{22.4}$ = 3.54mole

From the reaction expression;

6 moles of chlorine gas will react with 1 mole of P₄

3.54 mole of chlorine gas will completely react with $\frac{3.54}{6}$ = 0.59mole of P₄

Mass of P₄ = number of moles x molar mass

Molar mass of P₄ = 4 x 31 = 124g/mol

Mass of P₄ = 0.59 x 124 = 73.2g

8 0

3 years ago

A measure of how close a series of measurements are to one another.

ser-zykov [4K]

Answer: Precision is a measure of how close a series of measurements are to one another. Precise measurements are highly reproducible, even if the measurements are not near the correct value. Darts thrown at a dartboard are helpful in illustrating accuracy and precision

Explanation: a if this wrong u fault u gave no detail

7 0

2 years ago

A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What ar

Leona [35]

Answer:

Mole fraction of alcohols in liquid phase $x_1=0.2727\& x_2=0.7273$ .