It is on a plate boundary so there are a lot of volcanoes in that area. All the volcanoes form a "ring" around the plate boundary.
Answer:
1. 12.6 moles
2. 8.95 moles
3. 2A + 5B → 3C
4. 48 moles
Explanation:
1. 2Fe + 3Cl₂ → 2FeCl₃
We assume the chlorine in excess. Ratio is 2:2
2 moles of Fe, can produce 2 moles of chloride
12.6 moles of Fe will produce 12.6 moles of chloride.
2. 2Fe + 3Cl₂ → 2FeCl₃
For the same reaction, first of all we need to convert the mass to moles:
500 g . 1mol / 55.85 g = 8.95 mol
As ratio is 2:2, the moles we have are the same, that the produced
4. The reaction for the combustion is:
2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)
We assume the oxygen in excess.
Ratio is 2:6, so 2 mol of ethane produce 6 moles of water
Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles
Answer:
a) Ksp = 7.9x10⁻¹⁰
b) Solubility is 6.31x10⁻⁶M
Explanation:
a) InF₃ in water produce:
InF₃ ⇄ In⁺³ + 3F⁻
And Ksp is defined as:
Ksp = [In⁺³] [F⁻]³
4.0x10⁻²g / 100mL of InF₃ are:
4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M InF₃. </em>Thus:
[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³
[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻
Replacing these values in Ksp formula:
Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>
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b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:
7.9x10⁻¹⁰ = [x] [0.05 + 3x]³
Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L
Solving from x:
x = -0.018 → False solution, there is no negative concentrations.
x = 6.31x10⁻⁶M → Right answer.
Thus, <em>solubility is 6.31x10⁻⁶M</em>
Answer:
HCO₂/H₂O is not the acid-base conjugate pair.
Explanation:
<em>Acid and conjugate base pairs differ by an H+ ion.</em>
Neither HCO₂ nor H₂O has lost or gained protons.
The conjugate acid of H₂O is H₃O⁺
The conjugate base of HCO₃⁻ is CO₃²⁻
[A conjugate acid has one more H⁺ than its base]
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample