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Sedbober [7]
3 years ago
6

The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances

of 39K and 41k are 93.2581% and 6.7302%, respectively. Determine the isotopic mass of 41K.
Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
8 0

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:  

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu  

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 +  X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

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Answer:

Density = 2.7 \ g/cm^3

Explanation:

The equation for density is:

Density = \frac{mass}{volume}

We can plug the given values into the equation:

Density = \frac{81\ g}{30.0\ cm^2}

Density = 2.7 \ g/cm^3

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2 years ago
I am doing science, periodic table. what is an element that has 0 neutrons? I even searched online and it said that's not an ele
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Answer:

element hydrogen

Explanation:

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8 0
3 years ago
A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th
Ann [662]

Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

            Q = M x C x ΔT

            Q = -100 x 4.18 x (-5.14)

            Q = 2148.5 joules

            Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

          ΔH = Q / number of moles of NaHCO3

          ΔH = 2.1485 / 0.173

          ΔH = 12.42 KJ/mol

3 0
3 years ago
Đốt cháy hoàn toàn một amin đơn chức thu được CO2 và H2O theo tỉ lệ nCO2 : nH2O = 6 : 7. Vậy CT amin đó là
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7 0
3 years ago
He equation for the dissociation of pyridine is
sergeinik [125]

Answer:

10.10

Explanation:

Step 1: Write the basic dissociation reaction for pyridine

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq)      Kb = 1.9 × 10⁻⁹

Step 2: Calculate [OH⁻]

For a weak base, we will use the following expression.

[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M

Step 3: Calculate pOH

We will use the definition of pOH.

pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9

Step 4: Calculate pH

We will use the following expression.

pH = 14 - pOH = 14 - 3.9 = 10.10

3 0
3 years ago
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