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3 years ago

Explain why the first ionisation energy of beryllium is higher than the first ionisation

Chemistry

1 answer:

Debora [2.8K]3 years ago

6 0

Answer:

since the distance between the electron and the nucleus is smaller in B than in Li, the electrostatic nuclear force of attraction experienced by B is higher than the one experienced by Li...this translates to the higher first ionization energy in B than in Li.

Explanation:

i hope it's helping!

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Hey, please could someone answer this?

pav-90 [236]

Carbon dioxide is the other product created.

5 0

3 years ago

24. The air in a small 250 cm birthday balloon is at a pressure of 760 torr. A boy sits on it at

kvv77 [185]

Answer:

1.14atm

Explanation:

Given parameters:

V1 = 250cm³ ;

1000cm³ = 1dm³; so this is 0.25dm³

P1 = 760torr

760torr = 1atm

V2 = 220cm³ ; 0.22dm³

Unknown:

New pressure = ?

Solution:

To solve this problem, we apply Boyle's law and we use the expression below:

P1 V1 = P2V2

The unknown is P2;

1 x 0.25 = P2 x 0.22

P2 = 1.14atm

7 0

3 years ago

How many moles of oxygen must be placed

umka21 [38]

Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.

Further Explanation:

The Ideal Gas Equation is:

$PV = nRT$

where:

P - pressure (in atm)

V - volume (in L)

n - amount of gas (in moles)

R - universal gas constant $0.08206 \frac{L-atm}{mol-K}$

T - temperature (in K)

In the problem, we are given the values:

P = 2.00 atm (3 significant figures)

V = 3.00 L (3 significant figures)

n = ?

T = 25.0 degrees Celsius (3 significant figures)

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:

$Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15$

Therefore, for this problem,

$Temperature\ in \ K = 25.0 +273.15\\Temperature\ in \ K = 298.15$

Solving for n using the Ideal Gas Equation:

$n \ = \frac{PV}{RT}\\n \ = \frac{(2.00 \ atm) \ (3.00 \ L)}{(0.08206 \ \frac{L-atm}{mol-K})( 298.15 \ K)} \\n \ = 0.245 \ mol$

The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.

Learn More

1. Learn more about Boyle's Law brainly.com/question/1437490

2. Learn more about Charles' Law brainly.com/question/1421697

3. Learn more about Gay-Lussac's Law brainly.com/question/6534668

Keywords: Ideal Gas Law, Volume, Pressure

4 0

3 years ago

"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"

Artyom0805 [142]

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically, change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

4 0

3 years ago

A sample of Mo(NO3)6 has 2.22 x 10^22 nitrogen atom, how many oxygen atoms does the sample have?

goldfiish [28.3K]

There are 3.98 × 10^23 atoms of oxygen in the sample.

Given that;

1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen

x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen

x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms

x = 0.0368 moles

The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18

Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.

Learn more: brainly.com/question/9743981

4 0

2 years ago