Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
It is a combustion reaction because when methane burns with oxyzen it produces carbon dioxide,water and heat and light.
Answer:
Option B = 60,600 mg (correct option)
Explanation:
First of all we will have an idea which numbers are consider as significant.
1 = All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
2= Leading zeros are not consider as a significant figures. e.g. 0.02 in this number only one significant figure present which is 2.
3= Zero between the non zero digits are consider significant like 105 consist of three significant figures.
4= The zeros at the right side e.g 3400 are also significant. There are four significant figures are present.
In given options, Option A 60.6 mg have 3 significant figures.
Option B have 5 significant figures.
Option C have 4 significant figures.
Option D have 3 significant figures.
Thus option b is correct option which have more significant figures.
Answer:
hand picking
Explanation:
as stone is bigger in size we can see them with our eyes so we can handpick it
