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1 year ago

Use the periodic table to determine the electron configuration for dysprosium (Dy) and americium (Am) in noble-gas notation

Chemistry

2 answers:

sergij07 [2.7K]1 year ago

3 0

Answer:

Dysprosium [Dy]= $[Xe]4f^{10}6s^2$

Americium [Am]= $[Rn]5f^77s^2$

Dysprosium is a chemical element with symbol Dy and atomic number of 66. It is a rare earth metal and as it contains partially filled f sub shells, it belongs to f block. Xe is the nearest noble gas and has atomic number of 54.

Americium is a chemical element with symbol Am and atomic number of 95. It is a rare earth metal and as it contains half filled f sub shells, it belongs to f block. Radon is the nearest noble gas and has atomic number of 86.

Alex1 year ago

3 0

Dy: [Xe]6s24f10

Am: [Rn]7s25f 7 :)

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Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.

Lelu [443]

Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.

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1 year ago

Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K

Dmitry [639]

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

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2 years ago

What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?

Jlenok [28]

Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.

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1 year ago

Read 2 more answers

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Nikitich [7]

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1 year ago

You are asked to go into the lab and prepare an acetic acid - sodium acetate buffer solution with a ph of 4.00  0.02. what mola

Oxana [17]

Hello!

To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:

$pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )$

$\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75$

So, the mole ratio of CH₃COOH to CH₃COONa is 5,75

Have a nice day!

5 0

2 years ago