Answer:
1. C(2)+H2(1) -> C2H6(1)
2. NH3(2)+O2(3)-> HCN(2)+H2O(3)
I am not sure about the second one.
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Answer: 0.8M
Explanation:
Given that,
Amount of moles of NaCl (n) = ?
Mass of NaCl in grams = 1.40 g
For molar mass of NaCl, use the molar masses:
Sodium, Na = 23g;
Chlorine, Cl = 35.5g
NaCl = (23g + 35.5g)
= 58.5g/mol
Since, amount of moles = mass in grams / molar mass
n = 1.40g / 58.5g/mol
n = 0.024 mole
Now, given that:
Amount of moles of NaCl (n) = 0.024
Volume of NaCl solution (v) = 30.0mL
[Convert 30.0mL to liters
If 1000 mL = 1L
30.0mL = 30.0/1000 = 0.03L]
Concentration of NaCl solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 0.024 mole / 0.03 L
c = 0.8 M (0.8M means concentration is in moles per litres)
Thus, the concentration of the solution is 0.8M
Answer:
Depends how much water and the temperature of the water. To heat 1 mL of water by 1 degree C 1 cal of energy (4.184 Joules) is required. Assuming that the water is at 25 degrees C, to boil one litre (liter) of water you would require 75,000 cal or 313.8 kJ.
The birds could have more diseases and could pass it to other reptiles
