Answer: 9.9 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of combine with 1 mole of
Thus 0.33 mole of will combine with =
mole of
Thus is the limiting reagent as it limits the formation of product.
As 1 mole of give = 1 mole of 
Thus 0.33 moles of give =
of
Mass of 
Thus theoretical yield (g) of produced by the reaction is 9.9 grams
<u>Answer:</u> The mass of second isotope of indium is 114.904 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the mass of isotope 2 of indium be 'x'
Mass of isotope 1 = 112.904 amu
Percentage abundance of isotope 1 = 4.28 %
Fractional abundance of isotope 1 = 0.0428
Mass of isotope 2 = x amu
Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %
Fractional abundance of isotope 2 = 0.9572
Average atomic mass of indium = 114.818 amu
Putting values in equation 1, we get:
![114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu](https://tex.z-dn.net/?f=114.818%3D%5B%28112.904%5Ctimes%200.0428%29%2B%28x%5Ctimes%200.9572%29%5D%5C%5C%5C%5Cx%3D114.904amu)
Hence, the mass of second isotope of indium is 114.904 amu
Explanation:
When we add a non-volatile solute in a solvent then due to the impurity added to the solution there will occur an increase in the boiling point of the solution.
This increase in boiling point will be known as elevation in boiling point.
As one beaker contains seawater (water having NaCl) will have some impurity in it. So, more temperature is required by seawater to escape into the atmosphere.
Whereas another beaker has only pure water so it is able to easily escape into the atmosphere since, it contains no impurity.
Thus, we can conclude that level of pure water will decrease more due to non-volatile solute present in it as compared to seawater.
Answer:
Scientific Notation: 4.45963 x 109 km (29.811 A.U.) Scientific Notation: 4.53687 x 109 km (30.327 A.U.) By Comparison: If you weigh 100 pounds on Earth, you would weigh 110 pounds on Neptune.
