Answer: Na, S, Cl
Explanation:
Atomic size decreases as one moves from left to right on the periodic table with elements in the same period. This is as a result of the electrons increasing in the outer circle and thus being drawn to the protons in the nucleus which will lead to the outer shell area decreasing.
Sodium (Na) comes before Sulfur (S) which comes before Chlorine (S) so this is the decreasing order as they are all in the same period.
Answer:
The answer is 5 mol
Explanation:
Number Of Moles = Mass Of Substance ÷ Mass Of One Mole
Mass of NaOH = Molar Mass of Sodium + Oxygen + Hydrogen
23 + 16 + 1
40 g/mol
Number of Mole = 200/40 = 5 mol
Thus, The number of Moles in 200 grams of Sodium hydroxide is 5 mol
<u>-TheUnknownScientist 72</u>
Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:
</em>
<em>N₂(g) +O₂(g) ⇄ 2 NO(g)
</em>
<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
<em />
Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
![K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}](https://tex.z-dn.net/?f=K%3D7.70%3D%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%20%3D%5Cfrac%7B%282.8-2x%29%5E%7B2%7D%20%7D%7B%281.2%2Bx%29.x%7D)
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M
Answer:
electron cloud charge of manganese=-5
Explanation:
atomic number=number of elecron +neutron number
DATA
Neutron NO.=30
atomic NO.=25
e=?
A=e+n
25=e+30
e=25-30
e= -5
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
