<span>Temperature is defined as the rate at which molecules move or vibrate
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Data: molar mass 470 g/mol
Percent composition:
Hg = 85.0%
Cl = 15.0%
Solution:
1) Convert % to molar ratios
A. Base: 100 g
=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol
Cl = 15.0 g / 35.45 g/mol = 0.4231 mol
B. divide by the higher number and round to whole number
Hg = 0.4325 / 0.4231 = 1.00
Cl = 0.4231 / 0.4231 = 1.00
=> Empirical formula = Hg Cl
2) Find the mass of the empirical formula:
HgCl: 200.59 g/mol + 35.45 g/mol = 236.04
3) Determine how many times is the empirical mass contained in the molecular mass:
470 g/mol / 236.04 = 1.99 ≈ 2
=> Molecular formula = Hg2 Cl2.
Answers:
Empirical formula HgCl
Molecular Formula Hg2Cl2
Pressure of the gas P1 = 30.7 kpa
When it doubled P2 = 61.4 kpa
Temperature T1 = 0 => T1 =. 0 + 273 =273
Temperature T2 =?
We have pressure temperature equation P1T1 = P2T2
=> T2 = P1T1 / P2 = (30.7 x 273) / 61.4 = 136.5
So the temperature for doubling the pressure is 136.5.
Answer:
The answer to your question is 0.33 moles of H₃PO₄
Explanation:
Data
moles of Ca(OH)₂ = 0.050
moles of H₃PO₄ = ?
Process
1.- Write the balanced chemical equation
3Ca(OH)₂ + 2H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6H₂O
Reactants Elements Products
3 Ca 3
12 H 12
14 O 14
2.- Calculate the moles of phosphoric acid
3 moles of calcium hydroxide --------------- 2 moles of phosphoric acid
0.5 moles of calcium hydroxide ----------- x
x = (0.5 x 2)/3
x = 0.33 moles
Answer:
C. Rate = k[H2]^2[O2]
Explanation:
Rate law only cares about REACTANTS. Since, rate law can only be determined experimentally, I am assuming the given reaction mechanism is elementary reaction from which we can write the rate law.
Only H2 and O2 are part of rate law since they are reactants and also the coefficient in front of H2 goes as exponent on rate law to indicate the order of H2 in the reaction.
Rate= k [H2]^2 [O2]
