Mass of solute ( m1 ) = 50.0 g
mass of solvent ( m2 ) = 150.0 g
Therefore:
m/m = ( m1 / m1 + m2 )
m/m = ( 50.0 / 50.0 + 150.0 )
m/m = ( 50.0 / 200 )
m/m = 0.25
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.
Answer: is a reactant; it is present before the reaction occurs.
Explanation:
In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.
Thus and HCl are reactants here whereas , and are products.
Answer:
p-Nitro acetanilide. 2-Naphthol aniline dye.
Answer:
2.53×10²³ atoms
Explanation:
To solve this, convert grams to moles and moles to atoms.
To convert grams to moles, use the molar mass of C₁₂H₂₂O₁₁ (342.296 g/mol). Divide the mass by the molar mass to get moles. Moles = 0.4207
To convert moles to atoms, use Avogadro's number (6.022×10²³ atoms/mol). Multiply moles by Avogadro's number to get atoms. Atoms = 2.53×10²³