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Sonja [21]
2 years ago
5

what is the new pressure acting on a 2.5 l balloon if its original volume was 5.8 l at 3.7 ATM of pressure

Chemistry
1 answer:
Ivan2 years ago
8 0

Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 3.7 atm

P_2 = final pressure = ?

V_1 = initial volume = 5.8 L

V_2 = final volume = 2.5 L

Now put all the given values in the above equation, we get:

3.7atm\times 5.8L=P_2\times 2.5L

P_2=8.6atm

Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.

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2 years ago
A sample of gasoline contains various hydrocarbons, which comprise atoms of carbon, hydrogen, and oxygen. The hydrocarbons mix t
Naddika [18.5K]

Answer:

  • mixture
  • homogenous mixture  (of hydrocarbons)
  • compound

Explanation:

Mixture can be easily separated by physical methods. Homogeneity and heterogeneity of a mixture is determined by whether the components there in are in a single phase and evenly distributed or not.

A solution has a solute evenly dissolved in  solvent to form a liquid substance.

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4 0
2 years ago
Roundup, an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles of molecules are there in a 669.1-g sam
Vedmedyk [2.9K]

Answer:

2.4 ×10^24 molecules of the herbicide.

Explanation:

We must first obtain the molar mass of the compound as follows;

C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1

We know that one mole of a compound contains the Avogadro's number of molecules.

Hence;

169 g of the herbicide contains 6.02×10^23 molecules

Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.

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2 years ago
During photosynthesis, sunlight shining on a plant is absorbed. Through several chemical reactions, the plant produces sugar, a
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2 years ago
Read 2 more answers
Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:
irina1246 [14]

Answer:

0.5

Explanation:

2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)

Using ideal gas equation,

PV = nRT

28.7torr

Converting torr to atm,

= 0.0378atm

V = 0.597L

T = 27 °C

= 300 K

a) PV = nRT

(0.0378atm) * (0.597L) = n(0.0821) * (300k)

= 0.000915 mol

moles of water and chlorine = 0.000915 mol

From the above equation, the ratio of water to chlorine = 1 : 2

Therefore, mole of chlorine = 0.000915/2

= 0.000458 mol

mole fraction = moles of specie/moles of all the species present

= 0.000458/0.000915

= 0.5

5 0
2 years ago
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