Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:


We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

Now we can <u>put the values into the equation</u>:

Now we can <u>solve for the initial temperature of gold</u>, so:

ºC
I hope it helps!
Fluorite is harder than gypsum but softer than apatite. Thus, the correct option is B.
<h3>What is the hardness of any element?</h3>
The hardness of any element may be defined as the capability of a material to oppose the process of deformation and remains in actual shape precisely.
According to the table of hardness scales by Mohs, the increasing order of given hardness of given elements is as follows:
Gypsum < Fluorite < Apatite.
Therefore, Fluorite is harder than gypsum but softer than apatite. Thus, the correct option is B.
To learn more about the Hardness of elements, refer to the link:
brainly.com/question/23721736
#SPJ1
Answer:
Explanation:
<u>1) First law of thermodynamic (energy balance)</u>
- Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter
<u>2) Energy change of each substance:</u>
Heat released or absorbed = mass × Specific heat × change in temperature
- density of water: you may take 0.997 g/ ml as an average density for the water.
- mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g
- Specif heat of water: 1 cal / g°C
- Heat released by the hot water:
Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)
- Heat absorbed by the cold water:
Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)
- Heat absorbed by the calorimeter
Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)
<u>4) Balance</u>
49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)
Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K
Ccal = 23.6 cal/ K
- Convert to cal / K to Joule / K
23.6 cal / K × 4.18 J / cal = 98.6 J/K
Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.
Answer:
Rn] 5f14 6d4 7s2 mp: none d: none. Seaborgium ..... Element. Density. Boiling or Melting Point in °C. Electron Configuration. Symbol.
Explanation: