Answer:
A compound contains atoms of different elements chemically combined together in a fixed ratio. An element is a pure chemical substance made of same type of atom. ... Elements contain only one type of atom.
Explanation:
The graphics in the attachment is part of the question, which was incomplete.
Answer: Fr = 102N and angle of approximately 11°.
Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:
Fx = 70+40-10 = 100
Fy = 40-20 = 20
Now, as the forces form a triangle, the totalforce is:
Fr = 
Fr = 
Fr = ≈ 102N
To determine the angle requested, we use:
arctg H = 
arctg H = 
H = tg 0.2 ≈ 11°.
Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.
<h3>What is electrolysis?</h3>
Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.
The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.
Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium
0.1 F will discharge = 0.1/3 × 27 g of aluminium
mass of aluminium deposited = 0.9 g of aluminium.
Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.
Learn more about electrolysis at: brainly.com/question/26050361
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
Independent variable: the student chooses to miss class.
dependent variable: in result of the student missing class his grade goes down.
hyposthesis: if the student misses class multiple times, then his grades will go down.