Answer:- 1.90 atm
Solution:- It is based on combined gas law equation, PV = nRT
In this equation, P is pressure, V is volume, n is moles of gas, R is universal gas constant and T is kelvin temperature.
If we divide both sides by V then:

We know that, molarity is moles per liter. So, in the above equation we could replace
by molarity, M of the gas. The equation becomes:
P = MRT
T = 20 + 273 = 293 K
M = 
Let's plug in the values in the equation:
P = 
P = 1.90 atm
So, the pressure of the gas is 1.90 atm.
Answer:
in the periodic table we can see that the ions of Cl is greater than the ions in Na
128 ml is the voume of the balloon if the temperature of the gas increases to 320.0k.
Explanation:
given that:
T1 (initial temperature) = 300K
V1 ( initial volume) = 120ml
T2 (final temperature) = 320 K
V2 (final volume) = ?
Pressure remained constant throughout the process.
From the equation
= 
Since pressure is constant the equation will be:
= 
V2 = 
Putting the values in the above formula:
V2 = 
= 128 ml
128 ml is the volume of the gas if temperature increases from 3OO K to 320k
Answer:
water
Explanation:
The heat capacity of a material, along with its total mass and its temperature, tell us how much thermal energy is stored in a material. ... <u>The result is that the temperature of the water cube is much more stable than the air — the water changes much more slowly; it holds onto its temperature longer.</u>
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:

Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g


The percent yield of chloro-ethane in the reaction is 82.98%.