Answer:
Explanation:
sp² hybridization is found in those compounds having double bond .
Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .
Rest have sp³ hybridization because they are saturated compounds .
Answer:
a)M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88 g/mol
b) if 0.100g is used instead of 0.200g
M = 0.1 / 0.034 = 2.94 hence the molar mass will be too low
Explanation:
0.2000 gHZ gives 100ml acid solution
33.5 ml of 0.1025 M NaOH is required to prepare it
the moles = mass / molar mass
mass = 0.200 gHZ
moles = 0.0335*100 * 0.1025 = 0.034
therefore molar mass = mass / moles
M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88
if 0.100g is used instead of 0.200g
M = 0.1 / 0.034 = 2.94 hence the molar mass will be too low
Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
