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Alona [7]
3 years ago
8

Calculate the mass, in grams, of a single xenon atom (mxe = 131.29 amu ).

Chemistry
1 answer:
Mariana [72]3 years ago
5 0

Answer is 2.18017\times10^{-22}gms.

Explanation: To calculate the mass of 1 Xe-atom , we will require atomic mass of Xe which is 131.29amu.

This atomic mass is the mass of 1 mole of 1 Xe-atom. One mole of Xe-atom is 6.022\times10^{23} atoms of Xe. We will use this relation to convert Mass of Xe-atom into grams.

\frac{\text{mass of 1 atom}}{atom}=\frac{\text{mass of 1 mole of atom}}{6.022\times10^{23}atoms}

\text{mass of 1 atom}=\frac{\text{mass of 1 mole of atom}}{6.022\times10^{23}}

\text{mass of 1 Xe-atom}=\frac{131.29}{6.022\times10^{23}}

\text{mass of 1-Xe atom}=21.8017\times10^{-23}gms

                                               =2.18017\times10^{-22}gms


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dmitriy555 [2]

<u>Answer:</u> The fraction of atom's mass contributed by nucleus is 0.99

<u>Explanation:</u>

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

The isotopic symbol of Helium-4 atom is _2^4\textrm{He}

Number of electrons = 2

Number of protons = 2

Number of neutrons = 4 - 2 = 2

We are given:

Mass of He-4 atom = 6.64648\times 10^{-24}g

Mass of 1 electron = 9.10939\times 10^{-28}g

Calculating the mass contributed by the nucleus = m_{He}-2(m_e)

Mass of the nucleus of He-4 atom = 6.64648\times 10^{-24}-(2\times (9.10939\times 10^{-28}))=(6.64648-0.0018219)\times 10^{-24}=6.64466\times 10^{-24}

To calculate the fraction of atom's mass contributed by the nucleus, we use the equation:

\text{Fraction of atom's mass contributed by nucleus}=\frac{\text{Mass of nucleus}}{\text{Mass of atom}}

Putting values in above equation, we get:

\text{Fraction of atom's mass contributed by nucleus}=\frac{6.64466\times 10^{-24}g}{6.64648\times 10^{-24}g}=0.99

Hence, the fraction of atom's mass contributed by nucleus is 0.99

8 0
3 years ago
A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your
Mazyrski [523]

Answer:

pH= 1.17

Explanation:

The neutralization reaction between HBr (acid) and KOH (base) is given by the following equation:

HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)

According to this equation, 1 mol of HBr reacts with 1 mol of KOH. Then, the moles can be expressed as the product between the molarity of the acid/base solution (M) and the volume in liters (V). So, we calculate the moles of acid and base:

<u>Acid</u>:

M(HBr) = 0.15 M = 0.15 mol/L

V(HBr) = 50.0 mL x 1 L/1000 mL = 0.05 L

moles of HBr = M(HBr) x V(HBr) = 0.15 mol/L x 0.05 L = 7.5 x 10⁻³ moles HBr

<u>Base</u>:

M(KOH) = 0.25 M = 0.25 mol/L

V(HBr) = 13.0 mL x 1 L/1000 mL = 0.013 L

moles of HBr = M(HBr) x V(HBr) = 0.25 mol/L x 0.013 L = 3.25 x 10⁻³ moles KOH

Now, we have: 7.5 x 10⁻³ moles HBr > 3.25 x 10⁻³ moles KOH

HBr is a strong acid and KOH is a strong base, so they are completely dissociated in water: the acid produces H⁺ ions and the base produces OH⁻ ions. So, the difference between the moles of HBr and the moles of KOH is equal to the moles of remaining H⁺ ions after neutralization:

moles of H⁺ = 7.5 x 10⁻³ moles HBr - 3.25 x 10⁻³ moles KOH = 4.25 x 10⁻³ moles H⁺

From the definition of pH:

pH = -log [H⁺]

The concentration of H⁺ ions is calculated from the moles of H⁺ divided into the total volume:

total volume = V(HBr) + V(KOH) = 0.05 L + 0.013 L = 0.063 L

[H⁺] = (moles of H⁺)/(total volume) = 4.25 x 10⁻³ moles/0.063 L = 0.067 M

Finally, we calculate the pH after neutralization:

pH = -log [H⁺] = -log (0.067) = 1.17

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