Empirical formula is the simplest ratio of components making up the compound. the molecular formula is the actual ratio of components making up the compound.
the empirical formula is CH₂. We can find the mass of CH₂ one empirical unit and have to then find the number of empirical units in the molecular formula.
Mass of one empirical unit - CH₂ - 12 g/mol x 1 + 1 g/mol x 2 = 12 = 14 g
Molar mass of the compound is - 252 .5 g/mol
number of empirical units = molar mass / mass of empirical unit
= 
= 18 units
Therefore molecular formula is - 18 times the empirical formula
molecular formula - CH₂ x 18 = C₁₈H₃₆
molecular formula is C₁₈H₃₆
Answer:
See explanation
Explanation:
You see, we must cast our minds back to Charles' law. Charles' law gives the relationship between the volume of a gas and temperature of the gas.
Now, Micheal left the balloon outside at a particular temperature and volume the previous night. Overnight, the temperature dropped significantly and so must the volume of the gas in the balloon!
Remember that Charles' law states that, the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Since the pressure was held constant, the drop in the volume of gas in the balloon can be accounted for by the drop in temperature overnight.
The concentration of the original calcium ions is 0.005 M
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.
As such we have the equation;
Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)
Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol
= 0.0022 moles
Now;
1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by 0.0022 moles of CaCrO4.
Given that the volume of the solution is 0.440 L, the concentration of the solution is; 0.0022 moles/0.440 L
= 0.005 M
Learn more about molarity:brainly.com/question/8732513
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Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!
