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OLEGan [10]
3 years ago
6

You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca

n get the balloon's density below this value, it will float. You measure both the temperature in the room as well as the mass of the balloon you will use and find they are 23.5°C and 0.576 grams, respectively. What volume of balloon will generate a density equal to that of the air around it? (Hint: Use the relationship you calculated above to determine the air density at the temperature given)
Chemistry
1 answer:
defon3 years ago
5 0
The equation to be used are:

PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles

The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa

n = 0.576 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.019862 mol
(101,183.9 Pa)V = (0.019862 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.839×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.839×10⁻⁴ m³ * 1000
V = 0.484 L
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{ \sf{H _{2}  \: is \: excluded}}

Oxygen molecule loses one oxygen atom, and gains four hydrogen atoms, hence it is reduced.

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WHAT IS KNOWN AS HYBRIDIZATION​
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\red{ \bold {\huge{ \textit{ \boxed {{\blue{QUESTION}}}}}}}

WHAT IS KNOWN AS HYBRIDIZATION

\huge\boxed{\fcolorbox{red}{blue}{ ANSWER }}

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​What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water? 
dusya [7]

Answer:

The concentration of KBr is  C = 0.07899 \ mol   L^{-1}

Explanation:

From the question we are told that

      The mass of KBr is  m_{KBr}  = 2.35 \ g

       The molar mass of KBr is  M_{KBr} =  119 g/mol

       Volume of water is V = 250 \ mL = 250 *10^{-3} =  0.250 \ L

This implies that the volume of  the solution is  V = 250 mL

The number of moles of KBr is

         n = \frac{m_{KBr}}{M_{KBr}}

Substituting values

         n =  \frac{2.35}{119}

        n = 0.01975 \ mol

The concentration of KBr is mathematically represented as

                C = \frac{0.01975}{0.250}

                C = 0.07899 \ mol   L^{-1}

3 0
3 years ago
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