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:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Solute particles can be atoms, ions or molecules.
Explanation:
Solute is the material which has to be mixed in the solvent to prepare a solution. So the concentration of solute should be less than the solvent. Also the solute and solvent should be of same nature other they will not dissolve with each other. The solute can be made up of atoms, ions or molecules depending upon the solvent. If the solvent concentration is in moles, then the solute concentration can be taken as atoms, ions or molecules. Also the saturation point plays a main role in deciding the kind of particles taken for the solute.
Answer:
1. C + O₂ → CO₂
2. C + CO₂ → 2 CO
3. Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
