1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.
2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.
3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.
4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.
5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
The options for given question are as follow,
1) Methane molecules show hydrogen bonding.
<span>2) Ammonia molecules show hydrogen bonding. </span>
<span>3) Methane has stronger hydrogen bonding than ammonia. </span>
<span>4) Both the compounds do not show hydrogen bonding. </span>
<span>5) Both the compounds have strong hydrogen bonding.
</span>
Answer:
Correct answer is Option-2 (Ammonia molecules show hydrogen bonding).
Explanation:
Hydrogen bond interactions are formed when a partial positive hydrogen atom attached to most electronegative atom of one molecule interacts with the partial negative most electronegative element of another molecule. So, in Ammonia hydrogen gets partial positive charge as nitrogen is highly electronegative. While the C-H bond in Methane is non-polar and fails to form hydrogen bond interactions.
Answer:
A) t = 22.5 min and B) t = 29.94 min
Explanation:
Initial concentration, [A]₀ = 100
Final concentration = 100 -75 = 25
Time = 45 min
A) First order reaction
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[25] − ln[100] = - 45k
-1.386 = -45k
k = 0.0308 min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[50] − ln[100] = - 0.0308 * t
-0.693 = -0.0308 * t
t = 22.5 min
B) Zero Order
[A] = [A]₀ − kt
Using the values from the initial reaction and solving for k, we have;
25 = 100 - k(45)
-75 = -45k
k = 1.67 M min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
[A] = [A]₀ − kt
50 = 100 - (1.67)t
-50 = - 1.67t
t = 29.94 min
Answer:
a) Volume of vial= 9.626cm3
b) Mass of vial with water = 62.92 g
Explanation:
a) Mass of empty vial = 55.32 g
Mass of Vial + Hg = 185.56 g
Therefore,
Density of Hg = 13.53 g/cm3
b) Volume of water = volume of vial = 9.626 cm3
Density of water = 0.997 g/cm3
Answer: The nuclide symbol of X is 
Explanation:
The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.
The general equation representing alpha decay process is:
For the given equation :
As the atomic number and mass number must be equal on both sides of the nuclear equation:
Thus the nuclide symbol of X is
