Answer:
So ²³⁵UF₆ effuses 1.0043 times faster than ²³⁸UF₆
Explanation:
The rate of effusion of two gases A and B can be expressed by <em>Graham's law</em>:
Where M is the molar mass, and in this case A is ²³⁵UF₆ while B is ²³⁸UF₆.
So now we <u>calculate the molar mass of each mass</u>:
²³⁵UF₆ ⇒235 + 6*19 = 349 g/mol
²³⁸UF₆ ⇒238 + 6*19 = 352 g/mol
Putting the data in Graham's law:
= 1.0043
So ²³⁵UF₆ effuses 1.0043 times faster than ²³⁸UF₆.
<u>Answer:</u> The mass of glucose in final solution is 1.085 grams
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
......(1)
Given mass of glucose = 15.5 g
Molar mass of glucose = 180.2 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:
To calculate the molarity of the diluted solution, we use the equation:
where,
are the molarity and volume of the concentrated glucose solution
are the molarity and volume of diluted glucose solution
We are given:
Putting values in above equation, we get:
Now, calculating the mass of glucose by using equation 1, we get:
Molarity of glucose solution = 0.0602 M
Molar mass of glucose = 180.2 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:
Hence, the mass of glucose in final solution is 1.085 grams
moles CO₂ = 5.57.10⁻⁴
<h3>Further explanation </h3>A mole is a number of particles(atoms, molecules, ions) in a substance
Can be formulated :
0.7% percent change with 3.5g of plant matter
mass :
moles :