Answer:
You will have 19.9L of Cl2
Explanation:
We can solve this question using:
PV = nRT; V = nRT/P
<em>Where V is the volume of the gas</em>
<em>n the moles of Cl2</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is 273.15K assuming STP conditions</em>
<em>P is 1atm at STP</em>
The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:
63g * (1mol / 70.906g) = 0.8885 moles
Replacing:
V = 0.8885mol*0.082atmL/molK*273.15K/1atm
V = You will have 19.9L of Cl2
Answer:
- <u><em>1.12 liters</em></u>
Explanation:
<u>Calculating number of moles</u>
- Molar mass of O₂ = 32 g
- n = Given weight / Molar mass
- n = 1.6/32
- n = 0.05 moles
<u>At STP</u>
- One mole of O₂ occupies 22.4 L
- Therefore, 0.05 moles will occupy :
- 22.4 L x 0.05 = <u><em>1.12 L</em></u>
Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.
For the compound, 
the electronic configuration of the atoms, carbon and hydrogen are:
Carbon (atomic number=6): In ground state= 
In excited state: 
Hydrogen (atomic number=1): 
All the bonds in the compound is single bond(
-bond) that is they are formed by head on collision of the orbitals.
The structure of the compound is shown in the image.
The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.
In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in of hydrogen will overlap to the 2p^{3}-orbital of carbon.
Thus, the hybridization of Hydrogen is 
-hybridization and the hybridization of Carbon is 
-hybridization.
The hybridization of each atom is shown in the image.
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
