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3 years ago

PLS HELP THE QUESTION IS ON THE PICTURE

Chemistry

1 answer:

IceJOKER [234]3 years ago

3 0

Concepts used:

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: Number of formula units = number of moles * 6.022*10²³

number of moles of an element or a compound = given mass/molar mass

__________________________________________________________

003 - Number of CaH₂ formula units in 6.065 grams

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

004 - Mass of 6.34 * 10²⁴ formula units of NaBF₄

Number of Moles:

We mentioned this formula before:

Number of formula units = number of moles * 6.022*10²³

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles= 10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

005 - Number of moles in 9.78 * 10²¹ formula units of CeI₃

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles OR 0.016 moles

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When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g

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Answer: The mass of iron produced will be 77.6 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For FeO:

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol$

For aluminium:

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol$

The given chemical reaction follows:

$3FeO+2Al\rightarrow 3Fe+Al_2O_3$

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = $\frac{3}{2}\times 0.93=1.395mol$ of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = $\frac{3}{2}\times 0.93=1.395moles$ of iron metal

Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

$1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g$

Hence, the mass of iron produced will be 77.6 grams

4 0

3 years ago

A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO

labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

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3 years ago