Although the models are not provided, I was able to find them and the beakers with solid present in them are:
1C
2A
2C
3A
3C
This is determined by the fact that the beakers all have a piece of closely packed substance laying at the bottom. This closely packed lattice is characteristic of solid substances, and the fact that they exist in the solution in the solid states indicates that they are insoluble.
Answer:
1.70 g.cm⁻³
Solution:
Data Given;
Mass = 84.7 g
Volume = 49.6 cm³
Density = ?
Formula Used;
Density = Mass ÷ Volume
Putting values,
Density = 84.7 g ÷ 49.6 cm³
Density = 1.70 g.cm⁻³
8
It’s 8 bc I said it was 8 ;)
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
Answer:
IUPAC Rules for Alkane Nomenclature
Find and name the longest continuous carbon chain.
Identify and name groups attached to this chain.
Number the chain consecutively, starting at the end nearest a substituent group.
Designate the location of each substituent group by an appropriate number and name.
Explanation:
