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.
Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen ,
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon ,
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
Cc stands for cm cubed (cubic centimetre).
So
6cc=6 cm^3.