We cannot see it its so small
Also yes it matches your hypothesis
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
Answer:
13.53 kJ
Explanation:
The energy of a gas can be calculated by the equation:
E = (3/2)*n*R*T
Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.
E = (3/2)*3.5*8.314*310
E = 13,531.035 J
E = 13.53 kJ
STP means standard temperature and pressure which is equivalent to 273 K and 1 atm, respectively. Assuming ideal gas behavior, the solution for this problem is as follows:
PV = nRT
Solve for n,
n = RT/PV
n = (0.0821 L-atm/mol-K)(273 K)/(1 atm)(1×10⁵ L)
<em>n = 2.24×10⁻⁴ moles</em>
Answer:
A. 15859.2 L or 15900 L
B. 0.629 mol
Explanation:
At STP, one mole is equal to approximately 22.4 L
L or mL is volume, so you are attempting to solve for L or mL.
A.
708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)
B.
(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.
