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never [62]
1 year ago
12

Assertion : When the added electron goes to smaller n=3 quantum level, the electron-

Chemistry
1 answer:
Mademuasel [1]1 year ago
7 0

It is due to low volume of electron that causes large distance between two electrons.

When the added electron goes to smaller quantum level, the repulsive force between two electrons is less because the electron which is added occupies small space due to less volume so there is a large distance between two electrons that leads to less force of repulsion.

If the electron has more volume so it occupies more space and there will more repulsive force between two electrons so that's why we can say that the smaller volume of electron is the reason of less repulsive force.

brainly.com/question/24387646

You might be interested in
A liquid has a volume of 34.6 ml and a mass of 46.0 g. what is the density of the liquid?
Minchanka [31]

The density of a substance can simply be calculated by dividing the mass by the volume:

density = mass / volume

 

Therefore calculating for the density since mass and volume are given:

density = 46.0 g / 34.6 mL

density = 1.33 g / mL

5 0
2 years ago
A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg
Dafna1 [17]

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

8 0
2 years ago
495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F
I am Lyosha [343]

Answer:

<u><em>Total pressure of the flask is 2.8999 atm.</em></u>

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

                                              = 0.495 L (1 cm³ = 1 mL = 0.001 L)                                            

Volume of nitrogen (N2) gas =  877 cm3

                                               = 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

                         = 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature =  25 °C

T = (25°C + 273.15) K

    = 298.15 K

Pressure = 114.7 kPa

               = 114.700 Pa

Pressure (torr) = 114,700 / 101325

                        = 1.132 atm

Formula:

PV=nRT  <em>(ideal gas equation)</em>

P = pressure

V = volume

R (gas constnt)=  0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

<u>For Oxygen:</u>

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

  = 0.560 / 24.47

  = 0.0229 moles

<u>For Nitrogen:</u>

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

  = 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

  = 0.0229 moles + 0.0406 moles

n  = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K  /  0.536 L

P = 1.554 / 0.536

<u><em>P = 2.8999 atm</em></u>

Total pressure in the flask is  2.8999 atm, while assuming the temperature constant.

7 0
2 years ago
Read 2 more answers
A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
Llana [10]

<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
                                    = </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

4 0
2 years ago
The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

The density of O₂ gas is 1.71 \frac{g}{L}

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

  • P=1.27 atm
  • molar mass of O₂= 32 \frac{g}{mol}.
  • R= 0.0821 \frac{atm*L}{mol*K}
  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
2 years ago
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