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Molodets [167]
4 years ago
5

Which of the following is not a valid equation for describing the behavior of

Chemistry
2 answers:
bonufazy [111]4 years ago
5 0
D is the answer!!!!!!!!!!!!!!!!
AnnZ [28]4 years ago
3 0

Answer:

A. (P1/V1) = (P2/V2) is not a valid equation

D. (V2/T1) =V2/T2) is not a valid equation.  ( V1/T1) = (V2/T2) is a valid equation

Explanation:

Step 1: Data given

The gas law is p*V = nRT (this means <u>option B is a valid equatio</u>n)

If we compare 2 gases we can write this as:

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )

Step 2: When temperature and number of moles are constant

This means volume and pressure will change.

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )  OR p1¨V1 = p2*V2

The relationship for Boyle’s Law can be expressed as follows: p1V1 = p2V2, where p1 and V1 are the initial pressure and volume values, and p2 and V2 are the changed values of the pressure and volume of the gas.

When the initial pressure rises, the initial volume will decrease (and vice versa). For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.

For the formula (P1/V1) = (P2/V2) it would mean pressure and volume are directly proportional, this is not true. So<u> option A is not a valid equation. </u>

Step 3: When pressure and number of moles are constant

This means volume and temperature will change.

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )  OR V1/T1 = V2/T2

For the formula (V1/T1) =V2/T2) it would mean volume  and temperature are directly proportional. This<u> is a valid equation</u>

<u>Option D</u> would only be a valid equation if its (V1/T1) =V2/T2)

(V2/T1) =V2/T2) is not a valid equation. ( I don't know if this was a thypo or not).

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Suppose 13.6 g of barium nitrate is dissolved in 300. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final mol
Elis [28]

Answer:

The molarity of barium cation in the solution is 0.173 M

Explanation:

Step 1: The balanced equation

Ba(NO3)2(aq) + Na2CrO4 (aq) → BaCrO4(s) + 2NaNO3(aq)

Step 2: Data given

Mass of Barium nitrate = 13.6 grams

Volume of 0.40M sodium chromate = 300 mL

Step 3: Calculate moles of Ba(NO3)2

Moles = mass / molar mass

Moles = 13.6 grams / 261.34 g/mol

Moles = 0.052 moles

Step 4: Calculate moles of Na2CrO4

Moles = Molarity * Volume

Moles Na2CrO4 = 0.40 * 0.3L

Moles Na2CrO4 = 0.12 moles

Step 5: Calculate limiting reactant

Na2CrO4 is in excess so all of Ba(NO3)2 will be consumed and reacts to form BaCrO4(s) in the form Ba2+

Step 6: Calculate moles of Ba2+

n(Ba2+)=n(BaCrO4) =n(Ba(NO3)2 = 0.0520 moles

Step 7: Calculate molarity of Ba2+

C=n/v so C(Ba2+)=0.0520/0.300 = 0.173 M

The molarity of barium cation in the solution is 0.173 M

4 0
3 years ago
PLZ HELP WILL GIVE BRAINLIEST
Marrrta [24]

Answer:

The answer is A

Explanation:

I checked on a temperature converter calculator.

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What layer of the skin contains blood vessels, nerves and hair follicles?
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Answer:

The answer to your question is Dermis

Explanation:

Below the epidermis is the dermis. This is where our blood vessels, nerve endings, sweat glands, and hair follicles are.

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What is the equilibrium associated with ka3 for h3po4?
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Phosphoric acid has 3 pKa values (pKa1=2.1,  pKa2=6.9, pKa3= 12.4) and after 3 ionization it gives 3 types of ions at different pKa values: 

H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq)         pKₐ₁ 
<span>

</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq)       pKₐ₂


HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq)          pKₐ₃ 

The last equilibrium is associated with the highest pKa value (12.4) of phosphoric acid. There the last OH group will lose its hydrogen and hydrogen phosphate ion (HPO₄²⁻) turns into phosphate ion (PO₄³⁻).
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Your lab partner named this compound 3-methyl-4-n-propylhexane, but that is not correct.
loris [4]

<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

<u>Explanation:</u>

The IUPAC nomenclature of alkanes are given as follows:

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  • A suffix '-ane' is added at the end of the name.
  • If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

We are given:

An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms

The image of the given alkane is shown in the image below.

Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

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