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3 years ago

6

Three quantum numbers for an electron in a hydrogen atom in a certain state are

1 answer:

OLga [1]3 years ago

6 0

It is option A 4s cause n=4, l=1 this is 4s orbital

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Name two units for density? I need help!!!!

yaroslaw [1]

Answer:

Density is commonly expressed in units of grams per cubic centimetre. For example, the density of water is 1 gram per cubic centimetre, and Earth ’s density is 5.51 grams per cubic centimetre. Density can also be expressed as kilograms per cubic metre (in MKS or SI units). For example, the density of air is 1.2 kilograms per cubic metre.

Explanation:

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3 years ago

A chemist repeats the gold foil experiment but she uses foil made of aluminum(atomic number 79). How would you expect her result

Tanya [424]

Omg I wish I knew this one sorry

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3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago

Calculate the molar solubility of CaCO3 in 0.250M Na2CO3<br><br> Kps CaCO3 is 4.96x10-9

Yuliya22 [10]

Answer:

solubility is 1.984x10⁻⁹M

Explanation:

When CaCO₃ is in water, the equilibrium that occurs is:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹

If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:

[Ca²⁺] [0.250M] = 4.96x10⁻⁹

Assuming you are adding an amount of CaCO₃:

[X] [0.250 + X] = 4.96x10⁻⁹

<em>Where X is the amoun of CaCO₃ you can add, that means, solubility</em>

X² + 0.250X - 4.96x10⁻⁹ = 0

Solving for X:

X = -0.25M → False answer, there is no negative concentrations.

X = 1.984x10⁻⁹M.

That means, <em>solubility is 1.984x10⁻⁹M</em>

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3 years ago

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I need help for some of the question please,

LenKa [72]

Answer:

Number 6

Explanation:

A reflex action is an involuntary action and nearly instantaneous movement in response to a stimulus.

5 0

3 years ago

Read 2 more answers