Answer:
Average atomic mass of chlorine is 35.48 amu.
Explanation:
Given data:
Percent abundance of Cl-35 = 76%
Percent abundance of Cl-37 = 24%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (76×35)+(24×37) /100
Average atomic mass = 2660 + 888 / 100
Average atomic mass = 3548/ 100
Average atomic mass = 35.48 amu
Average atomic mass of chlorine is 35.48 amu.
The sample response given in the question is right.
To find the answer, we need to know more about the distance and displacement.
<h3>How distance differ from displacement?</h3>
- Displacement is the shortest distance between the initial and final points of a body.
- It is the change in position with a fixed direction.
- Displacement is a vector quantity and can be positive, negative or zero values.
- Distance is the length of actual path of the body between initial and final positions.
- It's a scalar quantity and it can be positive or zero.
- The magnitude of displacement is less than or equal to the distance travelled.
Thus, we can conclude that the given sample response is right.
Learn more about distance here:
brainly.com/question/28124225
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Answer:
See below
Explanation:
Ammonium 
is not a molecular ion because it is just a poly-atomic ion. A molecular ion has a "negative or positive charge" as a whole but the positive charge on here is not on the whole. So, it is a poly-atomic ion and not molecular ion.
Answer:
The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock.
Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.
The time elapsed will be log (L) atoms X half-life.
Explanation:
Answer:
mass of HNO₃ = 0.378 g
Explanation:
Normality = Molarity * number of equivalents
Molarity = Normality/number of equivalents
normality of HNO₃ = 0.30 N, Volume = 20 mL
HNO₃ ionizes in the following way:
HNO₃(aq) ----> H⁺ + NO₃⁻
Therefore, number of equivalents for HNO₃ is 1
molarity of HNO₃ = 0.30/1 =0.30 mol/dm³
Using the formula, molarity = number of moles/volume in liters
number of moles = molarity * volume
Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL
number of moles = 0.006 moles
From the formula, mass = number of moles * molar mass
molar mass of HNO₃ = 63.0 g/mol
mass = 0.006 * 63
mass of HNO₃ = 0.378 g
