Answer:
The volume of a gas approaches zero as the temperature approaches absolute zero.
Step-by-step explanation:
You may have done a <em>Charles' Law experiment</em> in the lab, in which you measured the volumes of a gas at various temperatures.
You plotted them on a graph, and perhaps you were asked to extrapolate the graph to lower temperatures.
Your graph probably looked something like the one below.
There is clearly an x-intercept at some low temperature.
Inference: The volume of a gas approaches zero as the temperature approaches absolute zero.
Answer:
The answer is
<h2>0.95 atm</h2>
Explanation:
To solve the question we use the following conversion
That's
1 mmHg 
0.0013 atm
So we have
If 1 mmHg 0.0013 atm
Then 732 mmHg will be
732 × 0.0013 atm
We have the final answer as
<h3>0.95 atm</h3>
Hope this helps you
Answer:
Explanation:
Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.
Charges on group one.
Hydrogen = +1
Lithium = +1
Sodium = +1
Potassium = +1
Rubidium = +1
Cesium = +1
Francium = +1
Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.
Charges on group two.
Beryllium = +2
Magnesium = +2
Calcium = +2
Strontium = +2
Barium= +2
Radium = +2
Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.
Charges on group 13.
Boron = +3
Aluminium = +3
Gallium = +3
Indium = +3
Thallium= +3
Group 13 elements are also shows +1 charge by losing one valance electron.
Answer:
pH = 2 and pOH = 12
Explanation:
Given [OH⁻] = 1 x 10⁻¹²M
pOH = -log[OH⁻] = -log(1 x 10⁻¹²) = - ( -12 ) = 12
pH + pOH = 14 => pH = 14 - pOH = 14 - 12 = 2
Answer:
71g (assuming experimental data)
Explanation:
The balanced equation for this reaction:
+ 
→ 
+ 
Molar mass of H2SO4 = 98.1 g/mol
molar mass of NaOH = 40g/mol
Molar mass of Na2SO4 = 142.04g/mol
⇒ 1 mole or 98.1g of H2SO4 will yield 1 ole of NaSO4; alternately, 2 moles or 49 ×2 = 80g of NaOH produces 1 mole of NaSO4.
<em>Therefore, limiting reactant is NaOH.</em>
Assuming actual experiment is 20g of NaOH,
1 mole - 40g
x moles - 20g = 
= 0.5 moles
⇒1 mole of Na2SO4 - 142.04g
∴ 0.5 moles = 142.04 × 0.5
<u>= 71.02g</u>
