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2 years ago

If 3.5 g of element X reacts with 10.5 g of element Y to form the compound XY.

Chemistry

1 answer:

irina [24]2 years ago

8 0

Answer:

The percentage by mass of element X is 25 %.

The percentage by mass of element Y is 75 %

Explanation:

X + Y ⇒ XY

3.5 g of element X + 10.5 g of element Y = 14g in total

⇒Element X : 3.5g / 14g = 0.25 ⇒ x 100 % = 25 %

⇒Element Y : 10.5 / 14 = 0.75 ⇒ x 100 % = 75 %

The percentage by mass of element X is 25 %.

The percentage by mass of element Y is 75 %

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How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:

WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol 3 mol 1 mol 1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 + 2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol (76 g/mol) 3 mol (71 g/mol)

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

76 g 213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

CS₂ : Cl₂

76 g : 213 g

1 g : 2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

2.8 g of Cl₂ ≅ 1 g of CS₂

10 g of Cl₂ ≅ ? g of CS₂

by doing cross multiplication

g of CS₂ = 10 x 1 / 2.8

g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

8 0

1 year ago

Consider the compounds 4-tert-butylcyclohexanol and 4-tert-butylcyclohexanone. using silica gel plates, which of the two compoun

grandymaker [24]

1- we know that 4-tert-butylcyclohexanol is more polar than 4-tert-butylcyclohexanone (where the alcohols in general are more polar than ketons due to the hydrogen bond)
2- during separation via chromatography (in this case) the more polar solute will dissolve easily in polar solvents, where like dissolves like.
3- So, 4-tert-butylcyclohexanol will dissolve in ethyl acetete (which is polar) more than 4-tert-butylcyclohexanone, i.e, will have much higher Rf.
4- And also 4-tert-butylcyclohexanone will dissolve in dichloromethane (which lower in polarity than ethyl acetate) more than 4-tert-butylhexanol, i.e, will have much higher Rf

3 0

2 years ago

Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

As Mass = density × volume

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

PV = nRt

V = nRT/P

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0

2 years ago

Diatomic hydrogen gas and diatomic nitrogen gas react spontaneously to form a gaseous product. Give the balanced chemical equati

alex41 [277]

Answer : $NH_{3}$

Explanation : When diatomic hydrogen gas and diatomic nitrogen gas reacts spontaneously it forms ammonia gas as the product.

we will have the following reactions; As all the reactants and products are in the gaseous state the subscript (g) is given to the molecules.

$H_{2}_{(g)} + N_{2}_{(g)} ----> NH_{3}_{(g)}$

But this is not a balanced equation, as on the right side of reactants we have 2 hydrogen atoms and 2 nitrogen atoms which gives the product which has one nitrogen and three hydrogen in the compound state.

So, balancing the equation, we get,

$3H_{2}_{(g)} + N_{2}_{(g)} ----> 2NH_{3}_{(g)}$

We multiplied 2 with the diatomic hydrogen molecule which resulted in production of 2 moles of ammonia as the product.

So, the complete balanced equation is;

$3H_{2}_{(g)} + N_{2}_{(g)} ----> 2NH_{3}_{(g)}$

5 0

2 years ago

Read 2 more answers

Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta

Leokris [45]

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

3 0

2 years ago