Calculate the number of molecules in 64.5 grams of Nitrogen monoxide, NO

Why does ice float in water?

MaRussiya [10]

Ice has a lower density than the density of water.

5 0

2 years ago

Please help ill give you points and brainliest please

Paul [167]

Answer:

Answers with detail are given below

Explanation:

1) Given data:

Mass of Rb₃Rn = 76.19 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 478.43 g/mol

Number of moles = 76.19 g/ 478.43 g/mol

Number of moles = 0.16 mol

2) Given data:

Mass of FrBi₂ = 120.02 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 640.96 g/mol

Number of moles = 120.02 g/640.96 g/mol

Number of moles = 0.19 mol

3) Given data:

Mass of Zn₂F₃ = 88.24 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 187.73 g/mol

Number of moles = 88.24 g/ 187.73 g/mol

Number of moles = 0.47 mol

4) Given data:

Number of moles of Sb₄Cl = 1.20 mol

Mass of Sb₄Cl = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 522.49 g/mol

Mass = Number of moles × molar mass

Mass = 1.20 mol × 522.49 g/mol

Mass = 626.99 g

3 0

2 years ago

True or false atoms with fewer then 4 outer elections lend elections?

Mamont248 [21]

The answer is false ....

3 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

2 years ago

Besides glycerol what 3 other molecules make up a triglyceride?

ICE Princess25 [194]

Triglyceride is a molecule constituted by one one molecule of glycerol and 3 molecules of fatty acids. The structure of a glycerol is like a letter "E", where the vertical line ( | ) is the glycerol and the three horizontal lines are long chains of organic acids, each with one COOH group, called fatty acids. The answer is that the other 3 molecules that make up a triglyceride are fatty acids.

6 0

2 years ago