Answer:
0.6941 mg
Explanation:
First we <u>calculate how many LiNO₃ moles there are</u>, using the <em>given concentration and volume</em>:
- 1.0 mL * 0.10 M = 0.10 mmol LiNO₃
As 1 mol of LiNO₃ contains 1 mol of Li,<em> in the problem solution there are 0.10 mmol of Li</em> (the only metallic ion present).
Now we<u> convert Li milimoles into miligrams</u>, using its <em>atomic mass</em>:
- 0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg
It starts at the bottom and then it goes up into the air as evaporation and that’s when the air gets cooler so gas is cooler and the liquid would be hotter
Answer:
C. 1 proton 3 neutrons
Explanation:
A nucleus is more stable if the ratio of the neutrons to protons is between 1:1 and 1:1.5.
Thus the ratios of neutrons to protons for the nuclei are as follows
A- 1:1
B- 1:1
C- 1:3
D- 1:2
Among these ratios, C is the greatest thus the nucleus is the least stable.
Answer:
Explanation:
From the given information:
The density of O₂ gas = 
here:
P = pressure of the O₂ gas = 310 bar
= 
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴
= 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:
where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:
After solving;
V = 0.1152 L
∴
= 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.
Answer:
When iron filings and sulphur powder are mixed and heated they undergo a chemical reaction and form ferrous sulphide (FeS). It is a new substance which has properties entirely different from Fe and S. Therefore, heating of mixture of iron and sulphur powder is a chemical change.
