Answer:
c) 0.080 M Al₂(SO₄)₃
Explanation:
Ion [SO₄²⁻] concentration of each solution is:
a) 0.075 M H₂SO₄: <em>[SO₄²⁻] = 0.075M</em>. Because 1 mole of H₂SO₄ contains 1 mole of SO₄²⁻
b) 0.15 M Na₂SO₄: <em>[SO₄²⁻] = 0.15M</em>. Also, 1 mole of Na₂SO₄ contains 1 mole of SO₄²⁻
c) 0.080 M Al₂(SO₄)₃ [SO₄²⁻] = 0.080Mₓ3 =<em> 0.240M</em>. Because 1 mole of Al₂(SO₄)₃ contains 3 moles of SO₄²⁻.
<h3>Thus, the soluion that has the greatest [SO₄²⁻] is 0.080 M Al₂(SO₄)₃</h3>
Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
1) mass composition
N: 30.45%
O: 69.55%
-----------
100.00%
2) molar composition
Divide each element by its atomic mass
N: 30.45 / 14.00 = 2.175 mol
O: 69.55 / 16.00 = 4.346875
4) Find the smallest molar proportion
Divide both by the smaller number
N: 2.175 / 2.175 = 1
O: 4.346875 / 2.175 = 1.999 = 2
5) Empirical formula: NO2
6) mass of the empirical formula
14.00 + 2 * 16.00 = 46.00 g
7) Find the number of moles of the gas using the equation pV = nRT
=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)
=> n = 0.01769 moles
8) Find molar mass
molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol
9) Find how many times the mass of the empirical formula is contained in the molar mass
92.14 / 46.00 = 2.00
10) Multiply the subscripts of the empirical formula by the number found in the previous step
=> N2O4
Answer: N2O4
Given question is incomplete. The complete question is as follows.
Balance the following equation:

Answer: The balanced chemical equation is as follows.

Explanation:
When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.
For example, 
Number of atoms on reactant side:
H = 5
P = 1
O = 6
Ca = 1
Number of atoms on product side:
H = 6
P = 2
O = 9
Ca = 1
In order to balance this equation, we will multiply
by 2 on reactant side and we will multiply
by 2 on product side. Hence, the balanced chemical equation is as follows.
