Answer:
(a) See below
(b) 103.935 °F; 102.235 °F
Explanation:
The equation relating the temperature to time is

1. Calculate the thermometer readings after 0.5 min and 1 min
(a) After 0.5 min

(b) After 1 min

2. Calculate the thermometer reading after 2.0 min
T₀ =106.321 °F
ΔT = 100 - 106.321 °F = -6.321 °F
t = t - 1, because the cooling starts 1 min late

3. Plot the temperature readings as a function of time.
The graphs are shown below.
Answer:
B.) Trigonal planar
Explanation:
This molecule has 3 bonds and no lone pairs. The angles are all 120° and the bonds are within the same plane. These molecules have the molecular shape of trigonal planar.
Answer:
d. 1.2 × 1024
Explanation:
From the equation of reaction
2H2 + O2= 2H2O
i.e 2mole(4g) of hydrogen requires 1 mole(32g) of oxygen to produce 2mole (2×6.02×10^23 molecules) of H2O= 1.2×20^24 molcules of water.
NB: 1 mole of H2O contains 6.02×10^23 molecules of H2O