*WILL GIVE BRAINLIEST*

A sample of N2 gas occupied 9.20 L at 21 degrees C and 0.959 atm. If the pressure is tuned to 1.15 atm at constant temperature w

Gwar [14]

Newly occupied volume is 7.672L by N2 gas.

<u>Explanation:- </u>

Given

V= 9.20 L, T(i) = T (f) = 294 K, P (i) = 0.959 atm, P (f) = 1.15 atm

To Find V (f)

Solution

According to the gas equation

PV= nRT

Where P=Pressure

V=Volume

n=No of moles

R=Universal gas constant

T=Temperature

Also we know the relation between pressure, volume and temperature as

PV/T = constant

$\frac{p(i) V(i) \times p(f) V(f)}{T(i) \times T(f)}$

$V(f)=[P(i) V(i) T(f)] \div[T(i) P(f)]$

Where P(i)V(i)/T(i) are the initial values of Pressure, Volume and Temperature.

P(f)V(f)/T(f) are the final values of Pressure, Temperature and Volume.

Substitute the known values in the equation we get

= $\frac{0.959 \times 9,2 \times 294}{294 \times 1.15}$

= $2574.94 \div 338.1$

= 7.672 L

Thus the final volume of the system is 7.672 L

5 0

4 years ago

Calculate the empirical formula for each of the following substances. (Express answer as a chemical formula) 1) 2.90 g of Ag and

exis [7]

Answer:

1) Ag3N

2)Na2S

3)NaHSO4

4) KNO3

Explanation:

We divide each mass by the element's relative atomic mass

1) 2.90/108-Ag, 0.125/14-N

0.027-Ag, 0.0089-N

Divide by the lowest ratio

0.027/0.0089-Ag, 0.0089/0.0089 N

3-Ag, 1-N

Empirical formula- Ag3N

2)2.22/23-Na, 1.55/32-S

0.097-Na, 0.048-S

Divide by the lowest ratio

0.097/0.048-Na, 0.048/0.048-S

2-Na, 1-S

Empirical formula- Na2S

3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O

0.09-Na, 0.09-H, 0.09-S,0.366-O

Divide by the lowest ratio

0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O

1-Na, 1-H, 1-S, 4-O

Empirical formula- NaHSO4

4)1.84/39, 0.657/14-N, 2.25/16-O

0.047-K, 0.047-N, 0.14-O

Divide through by the lowest ratio

0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O

1-K, 1-N, O-3

Empirical formula- KNO3

4 0

3 years ago

Bella observes that the gas-filled tire of her bicycle becomes a little deflated when

s2008m [1.1K]

Answer:

Answer is A - The low temperature outside lowers the volume of the gas according to Charles's law because this law describes how a gas will behave at constant pressure.

Explanation:

4 0

3 years ago

These models show the electron structures of two different nonmetal elements.

Soloha48 [4]

Answer: Option (D) is the correct answer.

Explanation:

Valence shell is the shell present on the outermost core of an atom and electrons present in the valence shell are known as valence electrons.

If an atom has completely filled valence shell then it means the atom is not reactive in nature because it is already stable.

But when an atom has less than eight electrons in its valence shell then it means to attain stability the atom will readily attract electrons towards itself.

As the given element 1 has 8 electrons in its valence shell. Hence, it is not reactive in nature but element 2 has 6 valence electrons. So, in order to attain stability element 2 will readily attract 2 electrons from a donor atom.

Thus, we can conclude that element 2 is more reactive because it does not have a full valence shell, so it will attract electrons.

4 0

4 years ago

A chemist decomposes several samples of water into hydrogen and oxygen and weighs or more correctly measures the mass of the hyd

MariettaO [177]

Answer:

A. Hydrogen and oxygen combines in a constant mole ratio in all the three water samples

B. Carbon and oxygen are combined in a constant mole ratio in all the three samples of carbon dioxide

C. All pure samples of a compound contain the same elements combined in the same proportion by mass

D. Since the same elements are always combining to form a certain compound, and the atoms of each element has its own mass, the atoms of each element combining to form the compound do so in a fixed proportion by mass.

Explanation:

A. molar mass of hydrogen = 1.0 g/mol; molar mass of oxygen = 16 g/mol

The molar ratio of elements in a compound gives the proportion each element is combined with the other by mass.

molar ratio of hydrogen to oxygen in the samples:

Sample 1: Hydrogen; 1.5/1 = 1.5: Oxygen; 12/16 = 0.75

molar ratio = 1.5/0.75 = 2 : 0.75./0.75 = 1

molar ratio = 2:1

Sample 2: Hydrogen; 2/1 = 2: Oxygen; 16/16 = 1

molar ratio = 2/1 = 2 : 1/1 = 1

molar ratio = 2:1

Sample 3: Hydrogen; 2.5/1 = 2.5: Oxygen; 20/16 = 1.25

molar ratio = 2.5/1.25 = 2 : 1.25/1.25 = 1

molar ratio = 2:1

Hydrogen and oxygen combines in a constant mole ratio in all the three water samples

B. molar mass of carbon = 12 g/mol; molar mass of oxygen = 16 g/mol

molar ratio of carbon to oxygen in the samples:

Sample 1: Carbon; 0.5/12 = 0.04 : Oxygen; 1.3/16 = 0.08

molar ratio = 0.04/0.04 = 1 : 0.08/0.04 = 2

molar ratio = 1:2

Sample 2: Carbon; 1.0/12 = 0.083 : Oxygen; 2.7/16 = 0.168

molar ratio = 0.083/0.083 = 1 : 0.168/0.083 = 2

molar ratio = 1:2

Sample 3: Carbon; 1.5/12 = 0.125 : Oxygen; 4.0/16 = 0.250

molar ratio = 0.125/0.125 = 1: 0.250/0.125 = 2

molar ratio = 1:2

Carbon and oxygen are combined in a constant mole ratio in all the three samples of carbon dioxide

C. All pure samples of a compound contain the same elements combined in the same proportion by mass

D. Since the same elements are always combining to form a certain compound, and the atoms of each element has its own mass, the atoms of each element combining to form the compound do so in a fixed proportion by mass.

3 0

3 years ago