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Dafna1 [17]
3 years ago
11

Under laboratory conditions of 25.0 degrees C and 99.5 kPa, what is the maximum number of liters of ammonia that could be produc

ed from 1.50 L of nitrogen according to the following equation?
N2(g) + 3H2(g) ---> 2NH3(g)
Chemistry
1 answer:
Minchanka [31]3 years ago
7 0

Answer:

Volume of ammonia 3 L.

Explanation:

Given data:

Temperature = 25°C ( 25+273= 298 k)

Pressure = 99.5 kpa (99.5/101 = 0.98 atm)

Volume of nitrogen = 1.50 L

Volume of ammonia = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Moles of nitrogen:

PV = nRT

n = PV/RT

n = 0.98 atm × 1.50 L / 0.0821 atm. L/mol. K × 298 K

n = 1.47 /24.5 /mol

n = 0.06 mol

Now we will compare the moles of nitrogen with ammonia.

                         N₂        :        NH₃

                          1          :          2

                           0.06   :        2×0.06 = 0.12 mol

Volume of ammonia:

PV = nRT

V = nRT/P

V = 0.12 mol× 0.0821 atm. L/mol. K × 298 K/  0.98 atm

V = 2.9 atm. L /0.98 atm

V = 3 L

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How many molecules of O2 are in 153.9 g O2?
MrRissso [65]

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

molar mass of O2 molecule = 16*2=32g/mol

n= mole

To find the mole

n= mass/ molar mass

n= 153.9/32

n=4.81mol.

To find the number of molecules of o

Nm= number of molecule

Nn = Number of mole

NA = number of Avogados

Nm= Nn * NA

Nm= 4.81 *6.02*10^23

Nm= 2.895*10^24

3 0
3 years ago
What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?
serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0
3 years ago
A compound of carbon and hydrogen having one double covalent bond between two of the carbon atoms in the molecule is known as wh
Alexandra [31]
It would be known as a meth_(insert - ane, - ene, and -yne) hydrocarbon. 1 bond = -ane 2 bonds = - ene And so on.
7 0
3 years ago
A syringe with 32.0 mL contains gas at 649.0 mmHg. The plunger is pushed in and now the volume is 19.5 mL. What is the new press
Natasha2012 [34]

Answer:

1.40 atm

Explanation:

P1V1 = P2V2

P2= P1* V1/V2

P1=649.0mmHg

V1= 32.0mL

V2= 19.5mL

P2=649.0mmHg*32.0ml/19.5ml=

1065.03mmHg

To get mmHg to atm. The formula is 1atm/760mmHg

Use the answer you got in mmHg to convert into (atm), If you were asked

1065.03mmHg*(1atm/760mmHg)= 1.40atm

6 0
3 years ago
Classify the two balanced equations representing two chemical reactions
Zinaida [17]

Single Replacement & Decomposition

<h3>Further explanation</h3>

Given

Two chemical reactions

1. Cl2 + 2NaBr ⇒2NaCl + Br2

2. 2NaCl ⇒2Na + Cl2

Required

Type of reaction

Solution

Equation 1 : A single replacement reaction is a chemical reaction in which one element replaces the other elements of a compound to produce new elements and compounds

General formula :

A + BC ⇒ AC + B

Equation 2 : the decomposition reaction of a compound into its constituent elements or compounds

General formula :

AB ⇒ A + B

4 0
3 years ago
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