Answer:
"A", "water changes from a gas to a solid to a liquid", according to this phase diagram, at at 0°C, as pressure is increased from 0atm to 10atm.
Explanation:
The question asks what happens at 0°C, as pressure is increased from 0atm to 10atm.
According to the question, the temperature is held constant. The pressure changes. In the phase diagram, we find the temperature 0°C on the horizontal axis, and all points where the temperature are 0°C are along that vertical line.
Since the pressure starts at 0atm and increases to 10atm, we start at the bottom, and move upward along that line, to see what phases of matter the substance changes to.
At the bottom, it is initially in a "gas" phase. As it moves up, it transitions to a "solid" phase. Later, as it continues moving up, it changes again into a "liquid" phase.
Thus, the answer would be "A", "water changes from a gas to a solid to a liquid", according to this phase diagram, at at 0°C, as pressure is increased from 0atm to 10atm.
Answer:
0.120M is the concentration of the solution
Explanation:
<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>
<em />
Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.
To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:
<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>
2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3
<em>Molar concentration:</em>
0.0300 moles NaNO3 / 0.250L =
<h3>0.120M is the concentration of the solution</h3>
Explanation:
Let us take the volume of block is x.
Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.
Also, buoyancy force 
= (volume submerged in water × density of water) + (volume in oil × density of oil)
= 
= 
g
As, W = V × density of graphite × g
It is given that density of graphite is 
or 2160 
.
So, W = 2160 V g
= (0.592 V \rho + 408 V) g - 2160 V g = 0
= 1752
= 2959.46 or 2.959 
is the density of oil.
It is given that mass of flask is 124.8 g.
Mass of 35.3 
oil = 
104.7 g
Hence, in second weighing total mass will be calculated as follows.
(124.8 + 104.7) g
= 229.27 g
Thus, we can conclude that in the second weighing mass is 229.27 g.
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
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