Answer:
The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules
Explanation:
The mass of ice in the beaker = 15.0 grams
The initial temperature of the ice = 0°C
The final temperature of the ice = 0°C
The latent heat of fusion of ice = 330 J/g
The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice
Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J
The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.
Answer:
24.32
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 24
Abundance (A%) = 78.70%
Isotope B
Mass of B = 25
Abundance (B%) = 10.13%
Isotope C:
Mass of C = 26
Abundance (C%) = 11.17%
Average atomic mass of Mg =..?
The average atomic mass of Mg can be obtained as illustrated below:
Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]
= 18.888 + 2.5325 + 2.9042
= 24.3247 ≈ 24.32
Therefore, the average atomic mass of magnesium (Mg) is 24.32
Answer:
2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO
Answer:
36s^5
Explanation:
We have;
M2X3 (s)------> 2M^3+(aq) + 3X^2-(aq)
If [M^3+(aq)] = [X^2-(aq)] = s
We then have;
Ksp = (2s)^2 * (3s)^3
Ksp = 4s^2 * 9s^3
Ksp = 36s^5
Note that Ksp is known as the solubility product. It is an equilibrum equation that shows the solubility of a solute in water.
Answer:
76,6 kg
Explanation:
A kg it's equal to 1x10^3 grams
A Gigagrams it's equal to 1x10^9 grams
Knowing this, a kg it's equal to 1x10^6 gigagrams
![7,66*10^{-5}[gigagram]*\frac{1*10^6 [kg]}{1 [gigagram]}= 76.6 [kg]](https://tex.z-dn.net/?f=7%2C66%2A10%5E%7B-5%7D%5Bgigagram%5D%2A%5Cfrac%7B1%2A10%5E6%20%5Bkg%5D%7D%7B1%20%5Bgigagram%5D%7D%3D%2076.6%20%5Bkg%5D)
