J. Because birds swim and fish don’t
The Average atomic weight of X is 28.7amu
Isotopes are atoms with the same number of protons but differing numbers of neutrons.
Different isotopes have various atomic masses.
The proportion of atoms with a particular atomic mass that can be found in a naturally occurring sample of an element is known as the relative abundance of an isotope.
An element's average atomic mass is computed as a weighted average by multiplying the relative abundances of its isotopes by their respective atomic masses, then adding the resulting products.
Using mass spectrometry, it is possible to determine the relative abundance of each isotope.
The atomic weight of the element will be a weighted average of the isotopes based on the relative abundance:
(27.730 x 0.6058) + (28.841 x 0.1835) + (31.321 x 0.2107) = 16.7988 + 5.2923+ 6.599 = 28.690 = 28.7 amu.
Average atomic weight of X is 28.7amu
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Answer:
(1) Chloroplast
Explanation:
Cells of living organisms are made up of certain function-specific structures called ORGANELLES. Some organelles are present in plant cells and absent in animal cells and vice versa. In a plant cell, one notable organelle that allows it perform the photosynthetic process is the CHLOROPLAST.
However, the chloroplast is predominantly found in the LEAF part of a plant. This is because leaf cells are the site of photosynthesis. Hence, according to this question, Joe would be able to tell whether the plant cell was from the leaf or the root by looking for CHLOROPLAST as a differentiating factor in each cell.
Answer : The number of moles of solute 
is, 0.0788 moles.
Explanation : Given,
Molarity = 0.225 M
Volume of solution = 0.350 L
Formula used:
Now put all the given values in this formula, we get:
Therefore, the number of moles of solute is, 0.0788 moles.
Answer:
Explanation:
Hello there!
In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:
Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:
Therefore, the partial pressure of chloroform turns out to be:
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