Answer:
Calculate the mass percent of a potassium nitrate solution when 15.0 g KNO3 is dissolved in 250 g
of water.
2. Calculate the mass percent of a sodium nitrate solution when 150.0 g NaNO3 is dissolved in 500 mL
of water. Hint: 1 mL water = 1 g water
3. Calculate the weight of table salt needed to make 670 grams of a 4.00% solution.
4. How many grams of solute are in 2,200 grams of a 7.00% solution?
5. How many grams of sodium chloride are needed to prepare 6,000 grams of a 20% solution?
Mass Percent = Grams of Solute
Grams of Solution X 100%
100%
Grams of Solute = Grams of Solution X Mass Percent
= 26.8 grams NaCl
= 670 grams X 4.00%
100%
100%
Grams of Solute = Grams of Solution X Mass Percent
= 154 grams solute
= 2,200 grams X 7.00%
100%
100%
Grams of Solute = Grams of Solution X Mass Percent
= 1,200 grams NaCl
= 6,000 grams X 20.0%
100%
Explanation:
Answer:
The disadvantages of each of the given model of electron configuration have been mentioned below:
1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.
2). Arrow and line diagrams make the counting of electrons and take up too much space.
3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.
Fluorine 20 (F - Atomic number 9 and atomic mass 20). Firstly we need to know what is beta decay. Beta decay occurs when one neutron changes into a proton and an electron therefore the atomic mass will remain the same as even though we loose a neutron it is replaced by a proton, the atomic number is always raised by 1 when one beta decay occurs. The produced electron is shot out of the nucleus at an incredible speed. This speedy electron we call a beta particle.
Ok now the reaction.
20 20 0
F -> Ne + e
9 10 -1
Remember the atomic number determines the nature of the element ( i.e what elemnt it is).
Hope this helps :).
<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
We are given that;
1 stadium holds = 1 × 10^5 people
Number of iron atoms is 1 × 10^18 atoms
Assuming the stadium would carry an equivalent number of atoms as people.
Then, 1 stadium will carry 1 × 10^5 atoms
Therefore,
To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.
Number of stadiums = Total number of atoms ÷ Number of atoms per stadium
= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium
= 1 × 10^13 Stadiums
Thus, 1 × 10^18 atoms would occupy 1 × 10^13 stadiums
N2(g) + 3 H2(g) = 2NH3(g)
Qc = (NH3^2) / { (N2)(H)^3)}
Qc= 0.48^2 /{ ( 0.60) (0.760^3) }= 0.875
Qc < Kc therefore the equilibrium will shift to the right. This implies that Nh3 concentration will increase
