A 100.0 mL sample of 1.020 M HCl is mixed with a 50.0 mL sample of 2.040 M NaOH in a Styrofoam cup. If both solutions were initi
1 answer:
Answer:
Explanation:
Given data:
0.1 L HCl × 1.020 mol/L = 0.102 mol HCL
0.05 L NaOH × 2.040 mol/L = 0.102 mol NaOH
NaOH + HCl \rightarrow H_2O + NaCl
0.102 mol HCl /1 mol HCl × 1 mol H2O = 0.102 mol H2O
0.102 mol H2O / 1 mol H2O × 57000 J = 5814 J
M(t)=150*1=150 g
we know that heat energy is given as
where c is specific heat
total energy released is = 57 \times 0.102 = 5.814 kJ
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Answer:
The maximum height covered is 3.25 m.
The horizontal distance covered is 9.81 m.
The total time in the air is 1.63 seconds.
Explanation:
The launch speed, .
Angle of launch with the horizontal,
So, the vertical component of the initial velocity,
.
The horizontal component of the initial velocity,
Let, t be the time of flight, to the horizontal distance covered
.
Not, applying the equation of motion in the vertical direction.
Where s is the displacement in time t, u is the initial velocity and a is the acceleration.
In this case, (from equation (i), s=0 (as the final height is same as the launch height) and
(negative sign is due to the downward direction).
seconds.
So, the total time in the air is 1.63 seconds.
From equation (i),
Total horizontal distance covered is
.
Now, for the maximum height, H, applying the equation of motion as
Here, v is the final velocity and v=0 (at the maximum height), and h=H.
So,
.
Hence, the maximum height covered is 3.25 m.