Change the places of 'acts against the motion of an object' and 'causes an object to change speed or direction'
This process is called filtration. This process is a solid-fluid separation by the use of a medium wherein only the fluid (gases or liquids) can go through it. The medium is called the filter while the fluid that passed through the filter is called the filtrate. The solid particles are the large particles which cannot pass through the filter.
Answer: 1.2642*10²⁵ on both sides
Explanation:
First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides
There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21
Now use Avogrado's constant
21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³
= 1.2642*10²⁵
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
<h2>Answer:</h2><h3>The temperature of the gas: V</h3>
The temperature of gas is a variable quantity. It can be changed by changing energy or pressure of gas.
<h3>The amount of gas in the tube (in terms of mass and moles): C</h3>
It is a constant entity. As mass of gas once taken can not be changed by changing temperature, pressure etc.
<h3>The radius of the tube: C</h3>
The radius of tube cannot change at any rate.
<h3>The temperature of the gas (changed by the water surrounding it): V</h3>
It can be changed by changing the temperature of water surrounding it.
<h3>The type of gas: C</h3>
It can never be changed.
<h3>The pressure of the gas: V</h3>
It can be changed by simply changing temperature and volume of gas.