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Oksana_A [137]
1 year ago
14

Calculate the molar mass of a 2.89 g gas at 346 ml, a temperature of 28.3 degrees Celsius, and a pressure of 760 mmHg.

Chemistry
1 answer:
malfutka [58]1 year ago
8 0

The molar mass of gas = 206.36 g/mol

<h3>Further explanation</h3>

In general, the gas equation can be written

\large{\boxed{\bold{PV=nRT}}}

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K

T = temperature, Kelvin

mass (m)= 2.89 g

volume(V) = 346 ml = 0.346 L

T = 28.3 C + 273 = 301.3 K

P = 760 mmHg=1 atm

The molar mass (M) :

\tt PV=\dfrac{m}{M}RT\\\\M=\dfrac{mRT}{PV}\\\\M=\dfrac{2.89\times 0.082\times 301.3}{1\times 0.346}\\\\M=206.36~g/mol

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Type in the correct values to correctly represent the valence electron configuration of oxygen: AsB2pC
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Answer:

2s²2p⁴

Explanation:

Oxygen is an element on the periodic table with a total of 8 electrons. It's electronic configuration is given as 2,6.

Using the orbital notation we write as 1s²2s²2p⁴

Also, the valence electrons are the electrons in the outermost shell of an atom. These electrons mostly determine the chemical properties of an atom.

Oxygen has a total of 6 electrons in its outermost shell and it is given as 2s²2p⁴

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2 years ago
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In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc
Hatshy [7]

0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

Kf of glucose = 1.86 °C

ΔT = (freezing point of solvent) - (freezing point of solution)

ΔT = 0.325 degree celsius

molality =?

ΔT = Kfm

rearranging the equation:

m = \frac{0.325}{1.86}

m= 0.17 M

molal concentration of the glucose solution is 0.17 M

3 0
1 year ago
A teacher cut an apple into three wedges of the same size. She dipped one wedge in lemon juice, dipped another in water, and lef
aleksandrvk [35]

Answer:

The responding variable of this experement is the outcome and that would be  that the one in lemon juice responded and the one  in water didn't (the other one is the control). Thus the responding varible is the one in lemon juice.

Explanation:

4 0
2 years ago
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A chemist uses 0.25 L of 2.00 M H2SO4 to completely neutralize a 2.00 L of solution of NaOH. The balanced chemical equation of t
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2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

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v₁=0.25 L
v₂=2.00 L
c₂-?

n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
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c₂v₂=2c₁v₁

c₂=2c₁v₁/v₂

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2 years ago
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A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration
Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= 100 ml

Initial moles of NO= \frac{2.60}{10^2} moles

Initial moles of H2= \frac{1.30}{10^2} moles

Concentration of NO at equilibrium= 0.161M

Concentration(in M)=\frac{moles}{volume(in litre)}

Moles of NO at equilibrium= 0.161(\frac{100}{1000})

                                            =\frac{1.61}{10^2} moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴\frac{2.60}{10^2}-x=\frac{1.61}{10^2}

⇒x=\frac{0.99}{10^2}

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<u>Equilibrium</u>:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

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⇒Kc=19.4760

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1 year ago
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