It's a thermodynamic quantity equal to the enthalpy minus the product of entropy and the absolute temperature
Answer:
Alkenes of the type R–CH=CH–R can exist as cis and trans isomers;
- The control group defined as a group that does not include any change to the inconstant being tested. Why is a control crucial in an experiment? The control group is crucial because it acts as a benchmark to variatethe results of the experiment to. The experimental group is the group that the scientist is examing . The experimental group receives a change to a variable, or the conditions allowed to change in the experiment.
- There are two types of control group : Positive control groups and Negative control groups
- The positive control group is a group that is contrive to produce the effect you are looking for in the experimental group. The positive control group shows the scientists that the craved results are achievable . This helps stop false negative outcome in the experimental group, where a negative result is receive but is due to a failure in the experiment instead than a truly negative result based on the experimental conditions.
- A negative control group is a group that is not exposed to the different scientist is testing, called the independent variable. A negative control group function as a benchmark to secure that the results that are got are actually due to the independent variable and not anything else
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Answer:
2.13 g
Explanation:
(142.04)(0.0150) = 2.13 g
Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L
