Carbon dioxide is created when hydrolic is mixed with calcium carbonate.
0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
Answer:
1. See explanation below
2. Density
3. Masses
Explanation:
1. Your picture is a bit too small to see the values but maybe this will help you.
To determine the maximum maximum mass in grams that triple beam balance can measure all you have to do is add up the maximum of each beam. So all you need to do is see the value at the last notch of each beam.
However, if you are referring to the picture that is attached in the bottom: The answer would be 610g. Because the last notches of each beam are as follows:
100 g
500 g
10 g
So we add that we get 610g.
2. density can be computed using the formula:
D = M/V
where:
D = density
M = mass
V = volume
As you can see in the both figures A and B measure 20 g, this means that their masses are the same. The density of objects can be different when either their masses, or their volumes are different. So even if they have the same mass, they can have different densities because they have different volumes.
3. Force of gravitational attraction between two objects is dependent on the masses of the two objects and the distance. The larger the mass, the stronger the gravitational force of attraction. This means that they have a direct relationship. Now when it comes to distance, the further apart they are the weaker the gravitational force of attraction, or in other words, they are indirectly related.
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A