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:
41.63g
Explanation:
Given parameters:
Volume of CaCl₂ = 500mL = 0.5L
Concentration = 0.75mol/L
Unknown:
Mass of the solute needed = ?
Solution:
The mass of the solute can be derived using the expression below;
Mass = number of moles x molar mass
But,
Number of moles = Concentration x Volume
So;
Mass = Concentration x Volume x molar mas
Molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol
Mass = 0.75 x 0.5 x 111 = 41.63g
First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
Attached is the balanced chemical equation xx
Answer:
hco3
Explanation: bc i said so
