Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234
So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
Molybdenum in periodic table
or
Molarity definition
1. LDFs
2. Intermolecular Forces
3. Intramolecular Forces
4. Linear
5. Tetrahedral
Edit: I'm new to this site and idk how to use it properly. I'm not sure about 2 and 3 currently because these forces are between molecules as well so INTERmolecular would be used twice (?)
Answer:
According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.
Explanation:
C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)
We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.
According to Le Chatelier's principle,
1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.
2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.
3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
