Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
To answer this lets first see how much 1 kg is equal to in cg.
1 kg = 100000 cg
Now lets multiply:-
100000 × 1.7 = <span>170000
</span>
So, 1.7 kg = <span>170000 cg
</span>
Hope I helped ya!!!
Well, if you look at group 1 of the periodic table, you will notice a thrend. All elements in group 1 have 1 valence / outer electron. Then you look at period 2, 3, 4 and so on, you will see that the group number corresponds the number of valence/ outershell electrons. Hence, the group determines the electron(s) on the outershell.
Answer:
~69.744 moles of Ca
Explanation:
Using Avogadro's constant , we know that:
1 mole = 6.022 x 10^23 atoms
S0, the number of moles in 4.20 x 10^25 atoms of Ca:
=(4.20 x 10^25 x 1 )/(6.022 x 10^23)
~69.744 moles of Ca
Q2:How many atoms are in 0.35 moles of oxygen?
1 mole = 6.022 x 10^23 atoms
S0, the number of atoms in 0.35 moles of oxygen:
=[0.35 x (6.022 x 10^23)]
=2.1077 x 10^23 atoms of Oxygen
Hope it helps:)
