Answer:
D
Explanation:
You need both, independant and dependant variables for a hypothesis.
Take note of the subscript written for each element in the compound. To find the total number of moles, make sure to multiply the subscript with the number of moles of compound. The answer for each is written below:
a. 3*1 = 3 moles Nitrogen; 3*3 = 9 moles Hydrogen
b. 0.25*2 = 0.5 moles Hydrogen; 0.25*1 = 0.25 moles Oxygen
c. 5*2 = 10 moles Hydrogen; 5*1 =5 moles Sulfur; 5*4 = 20 moles Oxygen
d. 0.75*1 = 0.75 moles Calcium; 0.75*1*2 = 1.5 moles Nitrogen; 0.75*3*2 = 4.5 moles Oxygen
Answer:
See notes on LeChatlier's Principle I gave you yesterday.
Explanation:
Remember chemical see-saw => Removing Fe⁺³ makes the reactant side of the see-saw lighter causing the balance to tilt right then shift left to establish a new equilibrium with new concentration values. Such would result in a decrease in FeSCN⁺² concentration and increases in Fe⁺³ and SCN⁻ concentrations to replace the original amount of ppt'd Fe⁺³. => Answer Choice 'B' ... Also, see attached => Concentration effects on stability of chemical equilibrium .
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol
5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2
H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol
We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.
</span> H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol
2.00 mol HF can be formed.
2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
The answer is elements in the first row (hydrogen and helium) will have outer electrons in the first energy level. Their principal quantum number is 1. Elements in the second row (lithium through neon) will have valence electrons in the second energy level with a principal quantum number of 2. I hope that helps.
