Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
Because the proton has a mass of 1 and a neutron has a mass of 1, we know that there is exactly one proton in the nucleus (because of the atomic number) which therefore tells us there are no neutrons as adding one would make the mass more than one.
Answer: m = 11.2 g C7H16
Explanation: First convert the mass of CO2 to moles. Then do the mole ratio between CO2 and C7H16 which is 7:1. Finally convert the moles of C7H16 to the mass of C7H16.
Solution attached.
<h3>
Answer:</h3>
83.33 seconds.
<h3>
Explanation:</h3>
<u>We are given;</u>
- Take off velocity as 300 km/hr
- Acceleration as 1 m/s²
We are required to calculate the take off time of the airplane.
<h3>Step 1: Convert velocity from km/hr to m/s </h3>
We are going to use the conversion factor.
The conversion factor is 3.6 km/hr per m/s
Therefore;
Velocity = 300 km/hr ÷ 3.6 km/hr per m/s
= 83.33 m/s
<h3>Step 2: Calculate the take off time</h3>
We know that;
v = u + at
where, u is the initial velocity, v the final velocity, a the acceleration and t is time.
But, initial velocity is Zero
Therefore;
83.33 m/s = 1 m/s² × t
Thus;
time = 83.33 m/s ÷ 1 m/s²
= 83.33 seconds
Therefore, the take off time is 83.33 seconds.
Hthe heat required to change the temperfature of 100 grams of water from 25 to 55 c is calculated as below
Q(heat) = M(mass) x C(specific heat capacity) x delta T(change in temperature)
M= 100 grams
C= 4.18 j/g/c
delta T= 55-25 =30c
Q=100 g x4.814 j/g/c x 30c = 12552 joules