answer: Surface water in nearby areas can become polluted.
Explanation:
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
Answer:
Option C. By increasing the temperature
Explanation:
From the graphical illustration above, we see clearly that the volume and temperature of the gas are directly proportional. This implies that as the temperature increases, the volume will also increase and as the temperature decreases, the volume will also decrease. This can further be explained by using the ideal gas equation as shown below:
PV = nRT
P is the pressure.
V is the volume.
n is the number of mole.
R is the gas constant.
T is the temperature.
PV = nRT
Divide both side by P
V = nRT/P
Since n and P are constant, the equation above becomes:
V & T
V = KT
K is the constant.
The above equation i.e V = KT implies that:
As T increases, V will also increase and as T decreases, V will also decrease.
Considering the question given above,
The volume of the gas can be increased if the temperature is increased.
Answer:
25,272
Explanation:
Multiply 9.0 x 108 = 972
Multiply 3.0 x 10 = 30 - 4 = 26
Lastly, multiply 972 x 26= 25,272
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
