Answer:
Option a. both homogeneous and heterogeneous mixture.
Explanation:
This can be explained as a mixture is the combination of more than one substances mixed together while retaining their identities or chemical properties.
- Homogeneous mixtures in which the components are mixed in the same amounts or proportion, i.e., it has uniformity in it.
- Heterogeneous mixtures in which the constituents are not mixed in uniform proportion and are separable.
- Pure elements are those which have the a stable isotope or the atomic number of which are same.
- Thus the composition can only vary in homogeneous and heterogeneous mixtures.
Just you think about it, a shower with hot water, you step out and it is cold, this happens because you have a enclosed space with the heat, so you get use to the heat, you step out it is cold. Hope this helped!
Answer: the molarity of the solution in volumetric flask "B' is 0.0100 M
Explanation:
Given that;
the Molarity of stock solution M₁ = 1.25M
The molarity os solution in volumetric flask A (M₂) = M₂
Volume of stock solution pipet out (V₁) = 5.00mL
Volume of solution in volumetric flask A V₂ = 25.00mL
using the dilution formula
M₁V₁ = M₂V₂
M₂ = M₁V₁ / V₂
WE SUBSTITUTE
M₂ = ( 1.25 × 5.00 ) / 25.00 mL
M₂ = 0.25 M
Now volume of solution pipet out from volumetric flask A V₂ = 2.00 mL
Molarity of solution in volumetric flask B (M₃) = M₃
Volume of solution in volumetric flask B V₃ = 50.00m L
Using dilution formula again
M₂V₂ = M₃V₃
M₃ = M₂V₂ / V₃
WE SUBSTITUTE
M₃ = ( 0.25 × 2.0) / 50.0
M₃ = 0.0100 M
Therefore the molarity of the solution in volumetric flask "B' is 0.0100 M
Answer:
d = 0.98 g/L
Explanation:
Given data:
Density of acetylene = ?
Pressure = 0.910 atm
Temperature = 20°C (20+273 = 293 K)
Solution:
Formula:
PM = dRT
R = general gas constant = 0.0821 atm.L/mol.K
M = molecular mass = 26.04 g/mol
0.910 atm × 26.04 g/mol = d × 0.0821 atm.L/mol.K×293 K
23.7 atm.g/mol = d × 24.1 atm.L/mol
d = 23.7 atm.g/mol / 24.1 atm.L/mol
d = 0.98 g/L
As,
CuCO₃ ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² = 1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x = 1.18 × 10⁻⁵ mol/L × 123.526 g/mol
x = 1.45 × 10⁻³ g/L
