Answer:
D) All of the above
Explanation:
When the temperature increases, it means that the kinetic energy of the gas molecules had increased too because the temperature is a way to measure this kinetic energy. When this happens, the velocity of gas molecules increases because kinetic energy and velocity are directed related:
Where <em>Ek</em> is the kinetic energy, <em>m</em> is the mass and <em>v</em> is the velocity.
If the velocity increases, there will be more collisions between molecules, so, with more collisions, the pressure, which is the measure of the force a gas makes in the system, will increase. So, all the answers explain why the pressure of a gas increases when the temperature increases.
Answer:
The new pressure of a gas initially at 575 mmhg and 12500ml and the volume changed to 15L is 479. 16 mmHg
Explanation:
Boyle's law states that the pressure of a given quantity of gas varies inversely with its volumes at constant temperature. It is represented as;
P V = K
P1 V1 = P2 V2
Where;
P1 = Initial pressure = 575 mmHg
P2 = Final pressure?
V1 = Initial volume = 12500ml = 12500÷ 1000 = 12.5 Liters
V2 = Final pressure = 15 Liters
Calculations :
P2 = P1 V1 ÷ V2
P2 = 575 × 12.5 ÷ 15
P2 = 7187 . 5 ÷ 15
P2 = 479. 16 mmHg
Therefore, the new pressure is 479. 16 mmHg
I've never done a reaction that involved just bond enthalpies before, so this is just a guess.
It's still products - reactants. The problem is the C=O bond. Do you count it once or twice? I'm going to choose once, but don't be surprised if it is incorrect.
Sum of the products - Sum of the reactants.
ΔH = 799 - 494
ΔH = 305
And that would be your answer to three places.
Because they have a full outer shell and they are both noble gases...Because the outer shell has attained stable electronics configuration of octate rule this it is with noble gas as helium.
Hope I helped!
Volume required for neutralization V will be:
V * 0.2125 M HCl = 25 mL * 0.17 M
V = 20 ml
First part:
When 10 mL is added we can apply Henderson equation to get the result, so:
The pH will be of basic buffer
pOH = pKb + log(salt/base)
or pOH = 4.19 + log (0.2125*10 / 25*0.17 - 10*0.2125 )
pOH = 4.19 and pH = 14 - 4.19 = 9.81
Second part:
When 20 ml is added, there is only salt formed
The pH will be salt of strong acid and weak base
So pH = 7 - 0.5 pKb - 0.5 log C
where C is the concentration of the salt formed so:
pH = 7 - (0.5*4.19) - (0.5 log (25*0.17) / (25+20))
= 5.42
Third part:
When 30 ml of the acid has been added,
The pH will be of the remaining strong acid
pH = - log (0.2125*10 / 25 + 30 )
= 1.326