Answer:
1. 25%
2. 1.25
3. 1
Explanation:
Be sure to look at the x and y axis to answer these questions. All you need to do is look at the graph.
- Hope that helps! Please let me know if you need further explanation.
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
<u>first step </u>
NO2(g) ------------------------------------> NO(g) + O(g)
<u>second step</u>
NO2(g) + O(g) -----------------------------> NO(g) + O2(g)
Explanation:
<u>first step </u>
NO2(g) ------------------------------------> NO(g) + O(g)
<u>second step</u>
NO2(g) + O(g) -----------------------------> NO(g) + O2(g)
Answer:
Fr
Explanation:
Francium (Fr) has the lowest ionization energy, with a value of 4.0727eV. This ionization energy increases as one advances in a period, being greater in the group of nonmetals.
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
