Answer:
I think the answer is 22.2
Explanation: What i DID was adding 12.0 + 10.0 and than that gave me 22 the I had added the to 0.200 and that how i got 22.2. Sorry if i got is wrong. :(
Answer:
Chemical, Physical, Chemical, Chemical, Physical!!
Explanation:
I just did it correctly.
The equilibrium vapour pressure is typically the pressure exerted by a liquid .... it is A FUNCTION of temperature...
Explanation:
By way of example, chemists and physicists habitually use
P
saturated vapour pressure
...where
P
SVP
is the vapour pressure exerted by liquid water. At
100
∘
C
,
P
SVP
=
1
⋅
a
t
m
. Why?
Well, because this is the normal boiling point of water: i.e. the conditions of pressure (i.e. here
1
⋅
a
t
m
) and temperature, here
100
∘
C
, at which the VAPOUR PRESSURE of the liquid is ONE ATMOSPHERE...and bubbles of vapour form directly in the liquid. As an undergraduate you should commit this definition, or your text definition, to memory...
At lower temperatures, water exerts a much lower vapour pressure...but these should often be used in calculations...especially when a gas is collected by water displacement. Tables of
saturated vapour pressure
are available.
V2 = 250 ml
Explanation:
Given:
P1 = 0.99 atm. V1 = 240 ml
P2 = 0.951 atm. V2 = ?
We can use Boyle's law to solve for V2
P1V1 = P2V2
V2 = (P1/P2)V1
= (0.99 atm/0.951 atm)(240 ml)
= 250. ml
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
