Answer:
See explanation below
Explanation:
First, you are not providing any data to solve this, so I'm gonna use some that I used a few days ago in the same question. Then, you can go and replace the data you have with the procedure here
The concentration of liquid sodium will be 8.5 MJ of energy, and I will assume that the temperature will not be increased more than 15 °C.
The expression to calculate the amount of energy is:
Q = m * cp * dT
Where: m: moles needed
cp: specific heat of the substance. The cp of liquid sodium reported is 30.8 J/ K mole
Replacing all the data in the above formula, and solving for m we have:
m = Q / cp * dT
dT is the increase of temperature. so 15 ° C is the same change for 15 K.
We also need to know that 1 MJ is 1x10^6 J,
so replacing all data:
m = 8.5 * 1x10^6 J / 30.8 J/K mole * 15 m = 18,398.27 moles
The molar mass of sodium is 22.95 g/mol so the mass is:
mass = 18,398.27 * 22.95 = 422,240.26 g or simply 422 kg rounded.
Answer:
1.67 atm.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm (P = ??? atm).
V is the volume of the gas in L (V = 5.0 L).
n is the no. of moles of the gas in mol (n = 0.5 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 203 K).
∴ P = nRT/V = (0.5 mol)(0.0821 L.atm/mol.K)(203 K)/(5.0 L) = 1.67 atm.
<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in = 
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
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