The molecular formula of organic solvent is <em>C6H12</em>
<h2>calculation</h2><h3>find the empirical formula first as in step 1 and 2</h3>
Step 1: f<em>ind the moles of C and H</em>
- moles = % composition/molar mass
- from periodic table molar mass of C= 12 g/mol while that of H= 1 g/mol
- moles is C is therefore = 85.6/12= 7. 13 moles
- moles of H= 14.4/1 - 14.4 moles
Step 2: <em>calculate the mole fraction by dividing each mole by smallest number of mole(7.13)</em>
H= 14.4/7.13 =2
the empirical formula is therefore = CH2
<h2>Then calculate the molecular formula from empirical formula</h2>
step 3: divide the grams molar mass by empirical formula mass
empirical formula mass = 12+(1 x2) = 14 g/mol
= 84.2/ 14 = 6
step 4: multiply each of the subscript within the empirical formula with the value gotten in step 3
- that is [CH2]6 = C6H12 therefore the molecular formula = <u>C6H12</u>
<span>1 mole of calcium carbonate reacts with 1 mole of sulfuric acid and produces 1 mole of calcium sulfate.
3.1660 g of CaCO3 is how many moles of calcium carbonate? 3.1660 / 100.0869 = 0.031633 moles.
3.2900 g of H2S04 is how many moles of sulfuric acid? 3.2900 / 98.079 = 0.033544 moles.
</span><span>The lesser of the two is 0.031633 moles.
Therefore, 0.031633 moles of calcium carbonate will combine with 0.031633 moles of sulfuric acid to produce 0.031633 moles of calcium sulfate.
Molecular weight of calcium sulfate is 136.14 g/mol.
Therefore, 0.031633 moles of calcium sulfate will weight 0.031633 x 136.14 g/mol = 4.3065 grams.</span>
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
All atoms are of <u>neutral charge, with an equal amount of protons and electrons.</u>
Answer:
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Explanation:
