Answer:
1. evaporation of ethanol
2. condensation of ethanol
Explanation:
The first and second process imply a change of phase, which is essentially a physical change because the matter change from liquid to gas and from gas to liquid respectively and it can ve reversed.
The options 3 to 6 are related to a chemical reaction because the outcomes of the process are different compounds than the ones we had at the beginning. The main reaction that implies process 3 to 6 is represented by:
C6H12O6 + impurities → 2 C2H5OH + 2 CO+ C
For option 5. burning of natural gas
, we have:
Natural gas(Mainly a mix of CH4 and C2H10) +O2→COn+H2O
Therefore, justo 1 and 2 are a physical change
Answer:Charge on X = (+ 2)
Explanation:
Charge on one Cl = -1
So Charge on two Cl = -2
The molecule = Neutral = 0 charge
So, to make total charge = 0 , the X should have +2 charge
XCl2 = +2 -2 = 0
X = +2
Alkaline earth metals(Be , Mg , Ca , Sr) form such type of compounds
Explanation:
<u>Answer:</u>
<em>The molarity of the
solution is
</em>
<em></em>
<u>Explanation:</u>
The Balanced chemical equation is
![1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)](https://tex.z-dn.net/?f=1AgNO_3%20%28aq%29%20%2B1KCl%20%28aq%29%20%3E%201%20AgCl%20%28s%29%2B1KNO_3%20%28aq%29)
Mole ratio of
: KCl is 1 : 1
So moles
= moles KCl
![Moles KCl = \frac {mass}{molarmass}](https://tex.z-dn.net/?f=Moles%20KCl%20%3D%20%5Cfrac%20%7Bmass%7D%7Bmolarmass%7D)
![= \frac {0.785 mg}{(39.1+35.5 g per mol)}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B0.785%20mg%7D%7B%2839.1%2B35.5%20g%20per%20mol%29%7D)
![= \frac {0.000785 g}{74.6 g per mol}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B0.000785%20g%7D%7B74.6%20g%20%20per%20mol%7D)
![= 0. 0000105 mol KCl](https://tex.z-dn.net/?f=%3D%200.%200000105%20mol%20KCl)
![= 0.0000105 mol AgNO_3](https://tex.z-dn.net/?f=%3D%200.0000105%20mol%20AgNO_3)
So Molarity
![= \frac {moles of solute}{(volume of solution in L)}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7Bmoles%20of%20solute%7D%7B%28volume%20of%20solution%20in%20L%29%7D)
![= \frac {0.0000105 mol}{26.2 mL}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B0.0000105%20mol%7D%7B26.2%20mL%7D)
= 0.000402M or mol/L is the Answer
(Or)
is the Answer
Measured values are verified and go under a series of experiments before being accepted as the norm. Accepted values are unanimously accepted values, seen as the norm by a qualified group of individuals (scientists or the people of the intellectual sphere).
1.22 litres will be the volume at a pressure of 1900 mm Hg if a gas occupies 3.06 liters at 760 mm Hg.
Explanation:
Data given:
Initial volume of the gas V1 = 3.06 Litres
Initial pressure of the gas P1 = 760 mm Hg
final pressure of the gas P2 = 1900 m Hg
final volume of the gas V2 =?
From the data we can see that Boyle's Law will be applied here,
P1V1 = P2V2
Rearranging the equation, we get
V2 = ![\frac{P1V1}{P2}](https://tex.z-dn.net/?f=%5Cfrac%7BP1V1%7D%7BP2%7D)
putting the values in the equation, we get
V2 = ![\frac{3.06 X 760}{1900}](https://tex.z-dn.net/?f=%5Cfrac%7B3.06%20X%20760%7D%7B1900%7D)
V2 = 1.224 Litres
The volume is 1.22 litres when the pressure is increased to 1900 mmHg.