<span>P*V/T=constant
so P*V= constant*T
if T doesn't change then
P*V= constant
so 150kPa*0.8L=75kPa*xL
xL=150kPa*0.8L/75kPa=1.6L
hope it help</span>
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Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is (
).
Utilize the titration method of
in view that we're given the concentrations of every compound and the quantity of
. Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.
- M1 = initial mass
- V1= initial volume
- M2 = final mass
- V2= final volume

- (0.138)(V1) = (0.205)x(26.0)
- V2=(0.205)x(26.0)\ 0.138
- V2 = 47.10 M/L
- The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L
Read more about the neutralization:
brainly.com/question/23008798
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Answer:
If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.
Explanation:
Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure
V ∝ 1/P or V₁·P₁ = V₂·P₂
Where:
V₁ = Initial volume
V₂ = Final volume = V₁/2
P₁ = Initial pressure = 832 torr
P₂ = Final pressure = Required
From V₁·P₁ = V₂·P₂ we have,
P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)
P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr