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DaniilM [7]
3 years ago
5

A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

what is the new volume of the balloon in the hot room? Use mc005-1.jpg.
Chemistry
2 answers:
Elina [12.6K]3 years ago
7 0

Using charles law, this would equal 3.68 L

belka [17]3 years ago
6 0

To solve this we assume that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = 313.15 x 3.50 / 298.15

<span>V2 = 3.68 L</span>

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If you start with 64g of a radioactive element how many half-lives would occur before 8g remain?
Nonamiya [84]

Answer:

3 half-lives

Explanation:

The half-life is the time that it takes to a radioactive element to decay to half of its initial amount.

Let's suppose we start with 64 g of the radioactive element.

  • After 1 half-life, the mass of the element will be 32 g.
  • After 2 half-lives, the mass of the element will be 16 g.
  • After 3 half-lives, the mass of the element will be 8 g.
3 0
3 years ago
Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -&gt; 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
3 years ago
What is the relationship between the number of each atom used to form compound and oxidation number?
deff fn [24]

Answer:Explanation:

In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species.

6 0
3 years ago
Each of the following names is wrong. Draw structures based on them, and correct the names:(b) 1,1,1-trimethylheptane
devlian [24]

The correct IUPAC name of 1,1,1-trimethylheptane is 2,2-dimethyloctane.

<h3>IUPAC NAME:</h3>

It is systematic way of nomenclature of organic compounds.

It is based on the position of functional groups, preference of the functional group, long or short chain of carbon, preference of double bond, single and triple bond, branching of carbon chain, etc.

The given compound is 1,1,1-trimethylheptane.

The given name is wrong according to IUPAC name because the numbering of carbon atom should be done in that way in which the carbon atom chain is largest.

Here, in this case the numbering is done from right side. Thus the largest carbon chain have 8 carbon atom.

If the numbering is done according to the question, number of carbon atom in straight chain is 7.

Thus, we concluded that the Correct IUPAC name of 1,1,1-trimethylheptane is 2,2-dimethyloctane.

learn more about Nomenclature:

brainly.com/question/14094007

#SPJ4

5 0
2 years ago
Identify the element: 6 energy level, metallic, 82 election
Lady bird [3.3K]

<u>Answer</u>:-

Name of element = Lead

Symbol ♾ of Lead = Pb

Atomic no. = 82

Atomic mass = 207.2 amu

No. of protons = 82

No. of electrons = 82

Yes, lead(Pb) is a metallic element and certainly it has 6 electron shells which means 6 energy level.

If you find anything unclear you can ask me...

6 0
3 years ago
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