To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.
For the first equation, we do a mass balance:
mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar
Assuming they have the same densities, then we can write this equation in terms of volume.
V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)
x + y = 150
For the second equation, we do a component balance:
1.00x + .13y = 150(.42)
x + .13y = 63
The two equations are
x + y = 150
x + .13y = 63
Solving for x and y,
x = 50
y = 100
Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.
Answer:
30.62 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 55 L
Initial pressure (P₁) = 3.2 atm
Initial temperature (T₁) = 520 K
Final temperature (T₂) = 760 K
Final pressure (P₂) = 8.4 atm
Final volume (V₂) =?
The final volume of the gas can be obtained as follow:
P₁V₁ / T₁ = P₂V₂ / T₂
3.2 × 55 / 520 = 8.4 × V₂ / 760
176 / 520 = 8.4 × V₂ / 760
Cross multiply
520 × 8.4 × V₂ = 176 × 760
4368 × V₂ = 133760
Divide both side by 4368
V₂ = 133760 / 4368
V₂ = 30.62 L
Therefore, the new volume of the gas is 30.62 L
Answer:
The second choice, or flammability.
Explanation:
The flammability of something is how easy it is for it to burn or ignite.
Answer:
Type of hybridisation
Explanation:
For example , saturated hydrocarbons have sp3 hybridisation while unsaturated hydrocarbons have either sp2 or sp hybridisation.
Example: Ethane( C2H6) has sp3 hybridisation and ethene(C2H4) has sp2 hybridisation and Ethyne(C2H2) has sp hybridization.