Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
Answer:
148 g
Explanation:
Step 1: Write the balanced equation for the decomposition of sodium azide
2 NaN₃ ⇒ 2 Na + 3 N₂
Step 2: Calculate the moles corresponding to 95.8 g of N₂
The molar mass of N₂ is 28.01 g/mol.
95.8 g × 1 mol/28.01 g = 3.42 mol
Step 3: Calculate the moles of NaN₃ needed to form 3.42 moles of N₂
The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.42 mol = 2.28 mol.
Step 4: Calculate the mass corresponding to 2.28 moles of NaN₃
The molar mass of NaN₃ is 65.01 g/mol.
2.28 mol × 65.01 g/mol = 148 g
speed and direction in which an object is moving. both speed and direction of motion. is a vector. two or more velocities add by velocity addition.
Answer:
375
Explanation:
325 + 50 = 375
just add them together and you get 375
The intermolecular bonding for HF is van der Waals, whereas for HCL, the intermolecular bonding is hydrogen. Since the van der Waals bond is stronger than hydrogen, HF will have a higher boiling temperature. Since the covalent bond is stronger than van der Waals, HF will have a higher boiling temperature.
