Answer:
Where is the diagram?please put the pictu
Answer:
THE PARTIAL PRESSURE OF OXYGEN GAS IN THE CONTAINER IS 92.67kPa WHICH IS OPTION B.
Explanation:
To calculate the partial pressure of oxygen gas collected over water, we use
Ptotal = Poxygen + P water
It is worthy to note that when oxygen is collected over water, it is mixed with water vapor and the total pressure in the container will be the sum of the pressure exerted by the oxygen gas and that of the water vapor at that given temperature.
At 20 C, the vapor pressure of water as given in the question is 2.33 kPa.
Using the above formula,
Ptotal = Poxygen + P water
Substituting for Poxygen, we have;
Poxygen = Ptotal - P water vapor
P oxygen = 95 .00 kPa - 2.33 kPa
P oxygen = 92.67 kPa.
The partial pressure of oxygen gas in the container is hence, 92.67kPa.
Explanation:
Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.
Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.
c = 
⇒ n
=
c
⋅ V
= 0.250 M ⋅ 6.00 L =
1.5 moles HCl
Now all you have to do is figure out what volume of 6.0 M stock solution will contain 1.5 moles of hydrochloric acid
c =
V = 
= 
=
0.25 L
Expressed in milliliters, the answer will be
→ rounded to two sig figs
Answer:
See explanation
Explanation:
Now we have, the graph attached.the stable disintegration product of C-14 is N-14.
Then;
Since the mass of C-14 originally present is 64g, at a time t= 17100 years, we will have;
N/No = (1/2)^t/t1/2
N = mass of C-14 at time t
No= mass C-14 originally present
t = time taken for N amount of C-14 to remain
No = mass of C-14 originally present
t1/2 = half life of C-14
N/64 = (1/2)^17,100/5730
N/64 = (1/2)^3
N/64 = 1/8
8N = 64
N = 8 g
