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Gre4nikov [31]
3 years ago
10

2. Calculate the atomic mass of an element that has two isotopes, each with 50.00% abundance. One isotope has a mass of 63.00 am

u and the other has a mass of 68.00 amu.
Chemistry
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

The atomic mass of element is 65.5 amu.

Explanation:

Given data:

Abundance of X-63 = 50.000%

Atomic mass of  X-63 = 63.00 amu

Atomic mass of X-68 = 68.00 amu

Atomic mass of element = ?

Solution:

Abundance of X-68 = 100-50 = 50%

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50×63)+(50×68) /100

Average atomic mass =  3150 + 3400 / 100

Average atomic mass  = 6550 / 100

Average atomic mass = 65.5 amu.

The atomic mass of element is 65.5 amu.

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My name is Ann [436]

Answer:

1.79 mol.

Explanation:

  • For the balanced reaction:

<em>2NaCl + F₂ → 2NaF + Cl₂. </em>

It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.

  • Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L (V = 18.3 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (299 K).

∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.

  • Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:

<em><u>Using cross multiplication:</u></em>

2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.

??? mol of NaCl is needed to react with → 0.895 mol of F₂.

∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.

4 0
3 years ago
The density of cadmium is 8.65 g/mL, what is the volume of a 33.4 g object made of pure cadmium?
Rzqust [24]

Answer:

Volume = 3.86 ml (Approx)

Explanation:

Given:

Density of cadmium = 8.65 g/ml

Mass of pure object = 33.4 g

Find:

Volume pure cadmium

Computation:

Volume = Mass / Density

Volume = 33.4 / 8.65

Volume = 3.86 ml (Approx)

6 0
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The 3 and 2 to the right of the components are subscriptions.

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Mazyrski [523]

Answer: half life

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k= disintegration constant

x= amount of substance left after time t

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3 0
3 years ago
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From the choices, the answers would be:
<span>they require only one electron to complete their outer shell
they have a high electronegativity</span>
6 0
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