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3 years ago

What is the mass of 5 moles of Fe2(CO3)3 ?

Chemistry

1 answer:

sineoko [7]3 years ago

7 0

Answer:

1218.585

Explanation:

Looking at the subscripts we know there are 2 atoms of Fe, 3 atoms of C, and 6 of O.

Take the molar mass of each atom (from the periodic table) and multiply by the # of atoms

Fe: 55.845×2= 111.69

C: 12.011×3= 36.033

O:15.999×6=95.994

Add the values together: 243.717 g/mol

That is 1 mole of the molecule. Multiply by 5 for the final answer.

243.717×5=1218.585

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Which segment represents melting?" Temperature (°C) Time (minutes)

zheka24 [161]

The segment that represents melting is time (minutes) and temperature.

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3 years ago

Gold has a molar (atomic) mass of 197 g/mol. consider a 2.47 g sample of pure gold vapor. (a) calculate the number of moles of g

adell [148]

N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol

n = 2.47 / 197
n = 0.01253 moles.
I'm sure you wanted to ask more than this. Just put some comments in. I can do the same.

3 0

3 years ago

A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place

kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8 cm³

mass of gold cube = 20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

7 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

How many moles of N2O5 are needed to produce 7.90 g of NO2?

Roman55 [17]

Answer:

0.085 moles of N₂O₅ are needed

Explanation:

Given data:

Mass of NO₂ produces = 7.90 g

Moles of N₂O₅ needed = ?

Solution:

2N₂O₅ → 4NO₂ + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

Number of moles = 7.90 g/ 46 g/mol

Number of moles = 0.17 mol

now we will compare the moles of NO₂ with N₂O₅.

NO₂ : N₂O₅

4 : 2

0.17 : 2/4×0.17 = 0.085 mol

Thus, 0.085 moles of N₂O₅ are needed.

4 0

3 years ago