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2 years 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]2 years 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.

Alex2 years ago

3 0

Dy: [Xe]6s24f10

Am: [Rn]7s25f 7 :)

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Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount

shtirl [24]

What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25 8.1
25/2 =12.5 + 8.1
12.5/2= 6.25 +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8 of its original mass.

(good luck on the regent if thats what your studying for :)

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2 years ago

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Motohiro felt chilled by the rainy afternoon. He decided to make a cup of hot tea. He let the tea leaves steep too long and the

kati45 [8]

Answer: the correct one would be option B.

Explanation:

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2 years ago

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.

Tema [17]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a: C $sp^{2}$ O $sp^{2}$ .

Explanation : The sigma bonds in the 'a' position involves carbon atom which undergoes $sp^{2}$ hybridisation and has $sp^{2}$ orbitals of carbon and as well as oxygen atoms involved in the bonding.

Answer 2) π Bond a: C has π orbitals and O also has π orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b: O $sp^{3}$ H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen and hydrogen atoms in it. It has $sp^{3}$ hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c: C $sp^{3}$ O $sp^{3}$

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of $sp^{3}$ hybridized bonds.

Answer 5) Bond d: C $sp^{3}$ C $sp^{3}$

Explanation : The bonding involved at the position of 'd' involves two carbon atoms. Therefore, they undergo $sp^{3}$ hybridization. The orbitals involved in this hybridization are also $sp^{3}$ .

Answer 6) Bond e: C1 containing O $sp^{2}$ C2 $sp^{3}$

Explanation : The bonding at the 'e' position involves two carbon atoms on is containing oxygen with double bonds and the C2 carbon atom. The carbon containing oxygen has $sp^{2}$ hybridized orbitals involved in the bonding; whereas C2 carbon ha $sp^{3}$ orbitals.

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2 years ago

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A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH

Olegator [25]

Answer:

0.0222 mole of NaOH is needed to react with NH4F

Explanation:

NH4F + NaOH --> NaF + NH3 + H2O

Data given

Mass of NH4F =0.821g, Concentration of NaOH= 1M, volume of NaoH =25ml

But mole = (CV)/1000

given mole of NaoH = (1 * 25)/1000 = 0.025moles of NaOH used

Molar mass of NH4F = 37g/mol

mole of NH4F used 0.821 / 37 = 0.0222 mole NH4F

Determine the excess and limiting reactant,

NaOH is in excess

0.025 - 0.0222 = 0.0028 mole NaOH excess

0.0222 mole of NaOH is required to react with NH4F

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2 years ago

A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Paul [167]

The difference of the sample mass and the heated sample mass is the amount of water which evaporated by heating process, then we can calculate:

mass of H₂O = 30.7 - 22.9 = 7.8 g

then calculate the number of moles:

number of moles of H₂O = mass / molecular mass

number of moles of H₂O = 7.8 / 18 = 0.43 moles

So the hydration number is 0.43.

Sr(NO₃)₂ • 0.43 H₂O

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2 years ago

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