You will have divided 50200 by 1000000 which 0.0502
Answer : The correct option is, (C) 2, 4 and 5.
Explanation :
Combustion reaction : It is a type of reaction in which a hydrocarbon react with an oxygen molecule to give carbon dioxide, water as a product.
For example : Methane react with oxygen to give carbon dioxide and water.
In the given list of chemical substances, 
are in oxide form. They can not be both reactant and product of a single combustion reaction.
In the given list, 
is the only hydrocarbon which shows a combustion reaction. That means react with 
to give 
and 
as a product.
The balanced combustion reaction of is,
Therefore, the correct answer is, (C) 2, 4, and 5.
Answer:
False
Explanation: sound travels fastest through solids slowest through liquids.
Answer:
Explanation:
<u>According to Arrhenius concept of acid and base:</u>
"When a base in a solution, produces/yields OH- (Hydroxide) ions."
So, when a base is dissolved in a solution, it produces OH- ions.
<u>For example:</u>
NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Answer:
0.581 L or 581 mL
Explanation:
As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)
Write down the amounts you are given.
V1 = 0.152 L (I was taught to always convert milliliters to liters)
P1 = 717 mmHg
T1 = 315 K
V2 = ?
P2 = 463 mmHg
T2 = 777 K
The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.
(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)
0.346 = (463*V2) / (777)
0.346*777 = (463*V2) / (777)*777
268.842 = 463*V2
268.842/463 = (463*V2)/463
V2 = 0.581
Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.
