If the molarmass of oxygen is 16 g/mol how many moles does that represent

Why do we need to use moles when we try to determine amounts of reactants and products in a reaction?.

$\huge\fbox{Answer ☘}$

Chemists use the mole unit to represent 6.022 × 10 23 things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro's number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units.

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8 0

3 years ago

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

sineoko [7]

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

7 0

3 years ago

Which type of wave does not require a medium to travel through?

nlexa [21]

Hello,

I'm not very sure but I think the correct answer is D) Water Wave

I really hope this helps!!!! Happy Holidays!!!! (:

8 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

What is common among all three types of forests in the Venn diagram?

Kruka [31]

Answer:

it looks like forests

Explanation:

8 0

3 years ago