Answer:
<h2>The answer is 70 kPa</h2>
Explanation:
In order to find the pressure if the volume were increased to 375 ml we use Boyle's law
That's

where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the final pressure

From the question
P1 = 105 kPa = 105 , 000 Pa
V1 = 250 mL
V2 = 375 mL
So we have

We have the final answer as
<h3>70 kPa</h3>
Hope this helps you
freezing is a phase change from liquid to solid
evaporation is a phase change from liquid to gas
sublimation is a phase change from solid to gas
condensation is a phase change from gas to liquid
MASHA ALLAH
Answer:
c,interpretation of the data
Explanation:
A good scientific explanation needs to be based on both data collected in an investigation and on the ability to interpret what the data means.
Normally, adding heat energy increases the temperature of a substance. However, during a phase change, the temperature of the substance stays constant. This happens because the heat energy is being used to change the phase of the matter instead of the temperature.
Answer:
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.
Explanation:
The unit of k is s⁻¹.
The order of the reaction = first order.
First order reaction: A first order reaction is a reaction in which the rate of reaction depends only the value of the concentration of the reactant.
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
[A] = the concentration of the reactant at time t
k= rate constant
t= time
Here k= 4.70×10⁻³ s⁻¹
t= 4.00
[A₀] = initial concentration of reactant = 0.700 M
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
![\Rightarrow -\frac{d[A]}{[A]}=kdt](https://tex.z-dn.net/?f=%5CRightarrow%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3Dkdt)
Integrating both sides
![\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt](https://tex.z-dn.net/?f=%5CRightarrow%5Cint%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3D%5Cint%20kdt)
⇒ -ln[A] = kt +c
When t=0 , [A] =[A₀]
-ln[A₀] = k.0 + c
⇒c= -ln[A₀]
Therefore
-ln[A] = kt - ln[A₀]
Putting the value of k, [A₀] and t
- ln[A] =4.70×10⁻³×4 -ln (0.70)
⇒-ln[A]= 0.375
⇒[A] = 0.686
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.