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.
According to the valence shell electron pair repulsion (VSPER) theory, an ammonia molecule <span> has a </span>trigonal pyramidal<span> shape with an experimental bond angle measure of 106.7 degrees. This is why it is difficult to accurately represent ammonia two-dimensionally because the molecular structure entails a 3-D projection with angles in it unlike the linear structure.</span>
Answer:
Relative and average atomic mass both describe properties of an element related to its different isotopes.
Explanation:However, relative atomic mass is a standardized number that's assumed to be correct under most circumstances, while average atomic mass is only true for a specific sample.
2.1648 kg of CH4 will generate 119341 KJ of energy.
Explanation:
Write down the values given in the question
CH4(g) +2 O2 → CO2(g) +2 H20 (g)
ΔH1 = - 802 kJ
2 H2O(g)→2 H2O(I)
ΔH2= -88 kJ
The overall chemical reaction is
CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ
CH4 +2 O2 → CO2 +2 H20
(1mol)+(2mol)→(1mol+2mol)
Methane (CH4) = 16 gm/mol
oxygen (O2) =32 gm/mol
Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O
which generate 882 KJ /mol
Therefore to produce 119341 KJ of energy
119341/882 = 135.3 mol
to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require
=135.3 *16
=2164.8 gm
=2.1648 kg of CH4
2.1648 kg of CH4 will generate 119341 KJ of energy
Answer:
Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.
However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.
