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3 years ago

How many moles are in 79.85g Fe2O3?

Chemistry

2 answers:

strojnjashka [21]3 years ago

8 0

Answer:

in 79.85 g of Fe2O3 there will be 0.5 moles of Fe2O3

Explanation:

To find the number of moles we must calculate the molecular weight of iron oxide, for that we use the periodic table:

mFe = 55.845 g/mol

mO = 15.999 g/mol

mFe2O3 = (2x55.845) + (3x15.999) = 159.687 g/mol

1 mole of iron oxide weighs 159.687 g, therefore, to calculate the number of moles present in 79.85 g it will be equal to:

nFe2O3 = 79.85/159.687 = 0.5 moles

mamaluj [8]3 years ago

4 0

Find molar mass of Fe2O3 = 159.7 g/mol
Then: 79.85g/159.7 g/mol = 0.5 mol

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3 years ago

Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The

Maksim231197 [3]

Answer:

$\large\boxed{\large\boxed{24.6kJ}}$

Explanation:

1. Energy to heat the liquid water from 55ºC to 100ºC

$Q=m\times C\times \Delta T$

m = 10.1g

C = 4.18g/JºC

ΔT = 100ºC - 55ºC = 45ºC

$Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J$

2. Energy to change the liquid to steam at 100ºC

$L=\lambda \times n$

λ = 40.6kJ/mol

n = 10.1g / 18.015g/mol = 0.5606mol

$L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J$

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3 years ago

What are the two fundamental portions of an atom?

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3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

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3 years ago

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Licemer1 [7]

Answer:

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Explanation:

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3 years ago