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:
C3H6 + Br2 → C3H6Br2
Explanation:
The reaction in which C3H6Br2 (1,2-Dibromopropane) is created is:
We can see that the only difference between the product (C3H6Br2) and the known reactant (C3H6) of the reaction is two bromine atoms (Br2). Br2 is diatomic bromine - a molecule we get after combining two bromine atoms. This compound is a red-brown liquid at room temperature, which means that that is the liquid described in your question.
Molybdenum in periodic table
or
Molarity definition
group 1 elements are metals with<u> low</u> density