Answer:
The Answer is 2,86 grs Na2CO3
Explanation:
What we have to do is find the mass of Na2CO3 as a pure component or solute. That's because the 11,8 mL are a solution of Na2CO3. This means, the sum between the solute Na2CO3 and water. To find the grams of Na2CO3 as pure component we create a factor series as is shown in the attached file.
Data:
Density of solution (ρ) = 1,10 grs sln Na2CO3/mL sln Na2CO3
Mass Percentage (%) = 22 grs Na2CO3/100 grs sln Na2CO3
The procedure is explained in the attached file
B is right yuh yuh yuh yuh
I think the answer is C. Hope this helped.
The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol at equilibrium.
The volume of the bottle = 4.7 L
Mass of ethanol = 0.33 g
Temperature (T1) = -11 oC = 273-11 = 262 K
P1 = 6.65 torr
Now we will calculate the mole by applying the ideal gas equation:-
PV = nRT
Or, n = PV/RT
Where P is the pressure
T is the temperature
R is the gas constant = 0.0821 L atm mol-1K-1
V is the volume
Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-
= 0.001913 mol C2H6
Conversion of the mole to gm
Molar mass of ethanol (M) = 46.07 g/mol
Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.
Hence, the mass of ethanol present in the vapor is found to be 8.8×10⁻²g.
Learn more about mole here:-brainly.com/question/15374113
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In thermodynamics<span>, </span>work<span> performed by a system is the energy transferred by the system to its surroundings. It can be calculated by the expression:
</span>
W = PdV
Integrating,
We will have,
W = P(V2 - V1)
133.7 (1 litre-atm / 101.325 Joule) ( <span>760 Torr / atm ) </span>= 783 (V2 - .0737 )
V2 = 1.35 L
Hope this answers the question. Have a nice day.
