Calculate the average atomic mas of an unknown element that has two naturally-occurring isotopes.

2 answers:

8 0

The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:

Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
</span>
Hope this helps.

jonny [76]2 years ago

7 0

Answer:

248.9 uma

Explanation:

We calculate the average atomic mass using the following formula

$m= \frac{(m_1.\%_1)+(m_2.\%_2)}{100} \\$

This formula depends on how many isotopes the atom has.

If the atom had three isotopes another term would be added to the formula

$m1= 248.7 uma\\\%_1= 75.47\%\\m_2= 249.4 uma\\\%_2= 24.53\%$

$m= \frac{(248.7)(75.47)+(249.4)(24.53)}{100}\\ m=248.9 uma$

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What is the molarity of a hydrochloric acid solution, when 30.0 mL is neutralized by 48.0 mL of 0.100 mol/L

bulgar [2K]

The molarity of a hydrochloric acid solution : 0.32 M

<h3>Further explanation </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution).

Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.

Acid-base titration formula

Ma. Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

1 ⇒HCl (valence=1, HCl ⇒H⁺+Cl⁻, one H⁺)

2⇒Ca(OH)₂(valence=2, Ca(OH)₂⇒Ca²⁺+2OH⁻, two OH⁻)

M₂=0.1 M

V₂=48 ml=0.048 L

V₁=30 ml=0.03 L

$\tt M_1.V_1.n_1=M_2.V_2.n_2\\\\M_1\times 0.03\times 1=0.1\times 0.048\times 2\\\\M_1=0.32$