The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of
<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>
This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas
<span><span>7014.04</span>=4.98≈5</span>
Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is
<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>
Answer:
Gravitational potential energy is energy an object possesses because of its position in a gravitational field.
Explanation:
1. 842g of NaOH will form 547.3 g of Al(OH)₃
2. The yield is 93.55%
<u>Explanation:</u>
3NaOH + Al → Al(OH)₃ + 3Na
1.
Molar mass of NaOH = 40 g/mol
Molar mass of Al = 27 g/mol
Molar mass of Al(OH)₃ = 78 g/mol
According to the balanced equation:
3 moles of NaOH requires 1 mole of Al to form 1 mole of Al(OH)₃
The ratio of NaOH : Al : Al(OH)₃ = 3 : 1 : 1
3 X 40 g of NaOH reacts with 27 g of Al to form 78 g of Al(OH)₃
120 g of NaOH + 27g of Al → 78 g of Al(OH)₃
120g of NaOH form 78g of Al(OH)₃
1g of NaOH will form g of Al(OH)₃
842g of NaOH will form of Al(OH)₃
= 547.3 g of Al(OH)₃
Therefore, 842g of NaOH will form 547.3 g of Al(OH)₃
2. Only 512 g of Al(OH)₃ is formed
Yield % = ?
Therefore, the yield is 93.55%
It’s a convergent boundary which means they are sliding
<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO