Answer:
- <em>Hydration number:</em> 4
Explanation:
<u>1) Mass of water in the hydrated compound</u>
Mass of water = Mass of the hydrated sample - mass of the dehydrated compound
Mass of water = 30.7 g - 22.9 g = 7.8 g
<u>2) Number of moles of water</u>
- Number of moles = mass in grams / molar mass
- molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol
- Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol
<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>
- The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g
- Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).
- Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol
<u>4) Ratio</u>
- 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂
Which means that the hydration number is 4.
Answer:
54
Explanation:
Given symbol of the element:
I⁻
Number of electrons found in an ion with the symbol:
This is a iodine ion:
For an atom of iodine:
Electrons = 53
Protons = 53
Neutrons = 74
An ion of iodine is one that has lost or gained electrons.
For this one, we have a negatively charged ion which implies that the number of electrons is 1 more than that of the protons.
So, number of electrons = 53 + 1 = 54
The number of electrons in this ion is 54
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The complete balanced chemical
equation is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6
mol H2O
First let us find for the limiting reactant:
>molar mass NH3 = 17 g/mol
moles NH3 = 54/17 = 3.18 mol NH3
This will react with 3.18*5/4 = 3.97 mol O2
>molar mass O2 = 32g/mol
moles O2 = 54/32 = 1.69 mol O2
We have insufficient O2 therefore this is the limiting
reactant
From the balanced equation:
For every 5.0 mol O2, we get 6.0 mol H2O, therefore
moles H2O formed = 1.69
mol O2 * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol
<span>mass H2O formed = 2.025*18 = 36.45 grams H2O produced</span>
