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rodikova [14]
3 years ago
12

What is the electron configuration for a neutral potassium atom

Chemistry
1 answer:
Feliz [49]3 years ago
6 0
1s2 2s2 2p6 3s2 3p6 4s1
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Lead: Pb, oxygen: O, and silver: Ag
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What is involved in weathering?
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A container of carbon dioxide has a volume of 260 cm at a temperature of 22.0°C. If the
Oxana [17]

Answer:

V₂ =279.4 cm³

Explanation:

Given data:

Initial volume = 260 cm³

Initial temperature = 22.0°C

Final temperature = 44.0°C

Final volume = ?

Solution;

22.0°C (22+ 273 = 295k)

44.0°C(44+273 = 317k)

Formula:

According to Charles's law

V₁/T₁ = V₂/T₂

Now we will put the values in formula:

V₂ = V₁×T₂ / T₁

V₂ = 260 cm³ × 317k / 295k

V₂ = 82420 cm³. k  / 295k

V₂ =279.4 cm³

3 0
4 years ago
Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibri
melisa1 [442]

Answer:

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

Explanation:

The equilibrium constant will remain the same irrespective of the amount of reactant taken.

Let us calculate the equilibrium constant of the reaction.

Kc=\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}

Let us calculate the moles of each present at equilibrium

moles=\frac{mass}{molarmass}

molar mass of PCl₅=208

molar mass of PCl₃=137

molar mass of Cl₂=71

moles of PCl₅ = \frac{mass}{molarmass}=\frac{4.13}{208}=0.02

moles of PCl₃= \frac{mass}{molarmass}=\frac{8.87}{137}=0.06

moles of Cl₂ = \frac{mass}{molarmass}=\frac{2.90}{71}=0.04

the volume is 5 L

So concentration will be moles per unit volume

Putting values

Kc = \frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}=0.024

Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

                PCl_{5}--->PCl_{3}+Cl_{2}

Initial                 0.02           0.06       0.04

Change             -x                   +x          +x

Equilibrium     0.02-x           0.06+x    0.04+x

Conc.                (0.02-x)/2       (0.06+x)/2   (0.04+x)/2

Putting values

0.024 = \frac{(0.06+x)(0.04+x)}{(0.02-x)2}

Solving

(0.024(2)(0.02-x)=(0.06+x)(0.04+x)

0.00096-0.048x=0.0024+x^{2}+0.1x

0.148x+x^{2}+0.00144=0

x = -0.01

so the new moles of

PCl₅ = 0.02 + 0.01  =0.03

PCl₃ = 0.06-0.01 = 0.05

Cl₂ = 0.04-0.01 = 0.03

mass of each will be:

mass= moles X molar mass

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

5 0
4 years ago
A chemist prepares a solution of silver(I) nitrate (AgNO3) by measuring out 269. micromol of silver(I) nitrate into a 300. mL vo
marishachu [46]

Answer:

The concentration of the chemist's silver(I) nitrate  solution is  0.897 mmol/L

Explanation:

<u>Step 1: </u>Data given

Number of moles AgNO3 = 269 micromol = 269 * 10^-6 mol

Volume of AgNO3 = 300 mL = 0.3 L

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2: </u>Calculate molarity of AgNO3

Molarity = number of moles per volume (in Liters)

Molarity AgNO3 = 269 *10^-6 mol / 0.3 L

Molarity AgNO3 = 8.97 * 10^-4 M

8.97 *10^-4 mol/L = 897 micromol/L = 0.897 mmol/L

The concentration of the chemist's silver(I) nitrate  solution is  0.897 mmol/L

8 0
3 years ago
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