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3 years ago

How many atoms are in 5.64 moles of iron

1 answer:

7 0

Hello devin ...The atomic mass of iron (Fe) is 55.85 amu. The atomic mass of radium (Ra) is 226 amu. One mole of atoms of any element contains 6.022 X 1023 atoms, regardless of the type of element. The mass of one mole of an element depends on what that element is, and is equal to the atom mass of that element in grams....i hope this helps you out

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How many ethyne molecules are contained in 84.3 grams of ethyne (C2H2)?

ololo11 [35]

(~26grams/mole) and Avogadros # (6.022x10^23) 84.3grams x 1mole/26grams x 6.022x10^23 molecules/mole = 1.95x10^24 molecules of C2H2

6 0

3 years ago

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Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO

Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$ $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$ $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

3 0

3 years ago

Consider the reaction. 2 Al ( s ) + Fe 2 O 3 ( s ) heat −−→ Al 2 O 3 ( s ) + 2 Fe ( l ) If 17.3 kg Al reacts with an excess of F

IrinaVladis [17]

Answer:

32.7 kilograms of aluminium oxide will be produced.

Explanation:

$2 Al ( s ) + Fe_2O_3 ( s ) +heat\rightarrow Al_2O_3 ( s ) + 2 Fe ( l )$

Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )

Moles of aluminium = $\frac{17300 g}{27 g/mol}=640.74 mol$

According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:

$\frac{1}{2}\times 640.74 mol= 320.37 mol$ of aluminum oxide

Mass of 320.37 moles of aluminum oxides:

320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg

32.7 kilograms of aluminium oxide will be produced.

6 0

3 years ago

Evaporation and condensation

bagirrra123 [75]

Answer:

Evaporation and condensation are two processes through which matter changes from one state to another.

Explanation:

7 0

3 years ago

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Choose all of the following mole ratios that are correct for the reaction given below. Be sure to choose all that apply.

Dennis_Churaev [7]

Answer:

2 moles NH3 = 1 mole of N2

3 moles of H2 = 1 mole of N2

3 moles of H2 = 2 moles of NH3

Explanation:

The balanced equation for the reaction is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Thus, we can say that:

1 mole of N2 = 3 moles of H2

1 mole of N2 = 2 moles of NH3

3 moles of H2 = 2 moles of NH3

Thus, considering the options given above, the right answers to the question are:

2 moles NH3 = 1 mole of N2

3 moles of H2 = 1 mole of N2

3 moles of H2 = 2 moles of NH3

5 0

2 years ago

Read 2 more answers