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IgorLugansk [536]
3 years ago
11

if a balloon had an initial volume of 100ml at pressure of 3.2 atm and the pressure was decreased to 1.9 atm, what would the new

volume of the balloon be assuming constant temperature
Chemistry
1 answer:
hram777 [196]3 years ago
4 0

Answer:

168.4 mL

Explanation:

Data Given

initial volume V1 of gas in balloon = 100 mL

initial pressure P1 of gas in balloon = 3.2 atm

final pressure P2 of gas in balloon = 1.9 atm

final volume V2 of gas in balloon = ?

Solution:

This problem will be solved by using Boyle's law equation at constant Temperature.

The formula used

                  P1V1 = P2V2

As we have to find out Volume, so rearrange the above equation

                  V2 = P1V1 / P2

Put value from the data given

V2 = 100 mL x 3.2 atm / 1.9 atm

V2 = 168.4 mL

So the final Volume of gas in baloon =  168.4 mL

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D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match th
Aleks04 [339]

Answer:

aldehyde

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ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

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6 0
2 years ago
CAN SOME ONE HELLLLLLLLLLLLLLLLLLLLLLP WILL GIVE BRAINLIEST TO THE BEST ANSWR FOR THIS AND GIVE AWAY 15 POINTS
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1. C
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I hope this helped! These are COMPLEX questions though! =D
8 0
3 years ago
A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o
azamat
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
                                                 = 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
                                               = 0.031395 g
                                               = 31.395 mg
                                               ≈ 31.4 mg
5 0
3 years ago
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