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3 years ago

Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.

1 answer:

7 0

1) Formulas:

a) mole fraction of component 1, X1

X1 = number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass

2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840

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Predict the products of La(s) + O2(aq) ->

mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

Hello there!

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

I hope this helps you to understand better. Enjoy your studies

3 0

2 years ago

When the temperature of a reacting mixture increases, the rate of reaction increases. Which statement explains why the rate of r

wlad13 [49]

Answer:

This is one of the factors that affects chemical reactions

Temperature:This is because when the temperature is raised energy in form of heat is supplied to the reacting particles and so the rate of reaction is increased.

3 0

3 years ago

Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure

o-na [289]

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

6 0

3 years ago

Nitric monoxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO₂), a dark brown gas. If 5.895 mol of NO is mixed with 2

Schach [20]

Answer:

Limiting reactant: O2

grams NO2 produced = 230.276 g NO2

grams of NO unused = 26.67 gNO

Explanation:

2NO + O2 --> 2NO2

Step 1: Determine the molar ratio NO:O2

molar ratio NO:O2 = 5.895: 2.503 = 2.35

stoichiometric molar ratio NO:O2 = 2:1

So, O2 is the limiting reactant.

Step2: Determine the grams of NO2:

?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2

Step 3: Determine the amount of excess reagent unreacted

moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted

moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted

mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted

8 0

3 years ago

According to the following reaction, how many grams of nitrogen monoxide will be formed upon the complete reaction of 31.0 grams

Mumz [18]

23.227 grams of NO is formed upon the complete reaction of 31.0 grams of oxygen gas with excess ammonia.

Explanation:

The balanced chemical equation is:

4NH3 + 5O2 ⇒ 4NO + 6H20

Number of moles will be calculated.

31 grams of oxygen is given

atomic weight of oxygen gas i.e O2 = 32gm/mole

The molar mass of oxygen gas is 32 grams/mole which is equal to one mole of the molecule.

Number of moles (n) = $\frac{mass}{atomic mass of one mole of the substance}$

n= $\frac{31}{32}$

n = 0.968 moles of oxygen are given for the reaction to occur.

From the balanced chemical equation it can be seen

5 moles of oxygen yielded 4 moles of NO

So, 0.968 moles of oxygen yield x moles of NO

$\frac{4}{5}$ = $\frac{x}{0.968 }$

4 × 0.968 =5x

x = 0.774 moles of NO will be formed.

To calculate the mass of NO

mass= atomic mass x number of moles

atomic mass of NO is 30.01 grams/mole

putting the values in the formula:

mass = 30.01 x 0.774

= 23.227 grams of NO is formed.

8 0

3 years ago