2. An element is a mixture of two isotopes. One isotope has an atomic mass of 34.969 amu and an
2 answers:
Answer: The average atomic mass of an element is 35.67 amu and the unknown element will be chlorine.
Explanation:
Mass of isotope 1 = 34.969 amu
% abundance of isotope 1 = 64.88% =
Mass of isotope 2 = 36.966 amu
% abundance of isotope 2 = (100-64.88)% =
Formula used for average atomic mass of an element :
Therefore, the average atomic mass of an element is 35.67 amu and the unknown element will be chlorine.
You might be interested in
The correct option that can be deduced for both Object P and Q is Option b) I and II only
To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.
<h3>What is Density?</h3>Density is a physical property of an object and can be expressed by using the relation:
From the given parameters, we are being told that:
- P → floats
- Q → sinks
This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.
However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.
Therefore, we can conclude that:
- 1. Q is heavier than P
- II. 1cm³ of Q has a greater mass than 1cm³ of P
Learn more about density here:
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = = -
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object