Answer:
when mass is 1×10⁴ Kg then density is 5 g/cm³.
when mass is 104 Kg then density is 5.2 × 10⁻² g/ cm³.
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
mass = 1×10⁴ Kg
volume= w ×l× h = 1×2× 1 = 2 m³
density = ?
first of all we will convert the given volume meter cube to cm³:
we know that
2×1000000 = 2 × 10⁶ cm³
Now we will convert the mass into gram.
1 Kg = 1000 g
1×10⁴ × 1000 = 1 ×10⁷ g
Now we will put the values in the formula,
d = m/v
d = 1 ×10⁷ g / 2×10⁶ cm³
d = 0.5 × 10¹ g/cm³
or
d = 5 g/cm³
If mas is 104 Kg:
104 × 1000 = 104000 g
d= m/v
d = 104000 g / 2×10⁶ cm³
d= 52000 ×10⁻⁶ g/ cm³
d= 5.2 × 10⁻² g/ cm³
Answer:
Explanation:
Given that:
Half life = 30 min
Where, k is rate constant
So,
The rate constant, k = 0.0231 min⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 0.0231 min⁻¹
Initial concentration = 7.50 mg
Final concentration = 0.25 mg
Time = ?
Applying in the above equation, we get that:-
Answer:
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? STP is a common abbreviation for "standard temperature and pressure." You have to recognize that five values are given in the problem and the sixth is an x. Also ... 273 1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr.
Explanation:
It has to be understood that 2 moles of oxygen are there in each mole of PbO2. Then it has to be calculated for 2 moles of oxygen.
Amount of oxygen = 2 * 5.43 moles
= 10.86 moles
Now it is also a fact that each mole of H2O contains 1 mole of oxygen. Then it can be easily concluded that 10.86 moles of water will be produced. I hope the procedure is clear enough for you to understand.
