Density of a solution is mass of solution per unit volume
Density = mass/volume
mass of solution is 46.08 g
volume of solution is 58.9 mL
since mass and volume is known, density can be calculated
density = 46.08 g / 58.9 mL = 0.78 g/mL
Answer:
2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)
Explanation:
This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:
HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)
However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:
2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)
Answer:
b) The dehydrated sample absorbed moisture after heating
Explanation:
a) Strong initial heating caused some of the hydrate sample to splatter out.
This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).
b) The dehydrated sample absorbed moisture after heating.
Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.
c) The amount of the hydrate sample used was too small.
It will create some errors but they do not create a difference of 13% difference as stated in the problem.
d) The crucible was not heated to constant mass before use.
Here the error is small.
e) Excess heating caused the dehydrated sample to decompose.
Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.
Answer:
Attached is the solution to the question
Explanation:
We first need to find the number of moles of gas in the container
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation
290 802.75 Pa x 5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol
therefore molar mass is 70.6 g/mol
