Answer: option C. Copper (II) chloride
Explanation:
To name CuCl2, we need to know the oxidation state of Cu in the compound as chlorine always have oxidation on —1 in all its compound. The oxidation state of Cu can be calculated as follows:
Cu + 2Cl = 0 (since the compound has no charge)
Cl = —1
Cu + 2(—1) = 0
Cu —2 = 0
Collect like terms
Cu = 0 +2
Cu = +2
Therefore, the oxidation state of Cu in CuCl2 is +2.
The name of the compound will be copper(ii) chloride, since cupper has oxidation state +2 in the compound.
Answer:
118750 ml
Explanation:
The chemical equation for complete combustion of hexane is given as;
2C6H14 + 19O2 → 12CO2 + 14H2O
From the equation of the reaction;
2 mol of C6H14 reacts with 19 mol of O2
2 ml of C6H14 reacts with 19 ml of O2
2500 mL of C6H14 would react with x ml of O2
2 = 19
2500 = x
x = 2500 * 19 / 2 = 23750 ml
Since oxygen is 20% of air;
23750 = 20 / 100 * (Volume of air)
Volume of air = 23750 * 100 / 20 = 118750 ml
Answer:
315.
Explanation:
Hello.
In this case, since the given number has five significant figures as the zero is to the right of the first nonzero digit (3), if it is required to report it with three significant figures, it is necessary to "cut" it at the first five without any rounding since the subsequent zero is less than five.
Thus the number turns out:
315
Best regards.
6.4 times 10^-7 is the answer because you are moving your decimal places to the left so it will be negative 7
Answer: 
Explanation:
Geometrical symmetry of the molecule and the polarity of the bonds determine the polarity of the molecule.
The molecule that has zero dipole moment that means it is a geometrically symmetric molecule and the molecule which has some net dipole moment means it is a geometrically asymmetric molecule.
As the molecule is symmetric, the dipole moment will be zero as dipole moments cancel each other and the molecule will be non-polar.
As the molecule is asymmetric, the dipole moment will not be zero and the molecule will be polar.
Example: 
Thus, we can say that
is a polar molecule.