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2 years ago

4. Why does the metallic character of the alkaline earth metals increase as you go down the

1 answer:

4 0

As you proceed down the periodic table, the metallic character becomes stronger. This is because as the atomic radius increases, there is less attraction between the nucleus and the valence electrons due to the greater distance between them, making electrons simpler to shed.

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A scientist finds an unusual object and decides to study it. Which observation indicates that the object is a living thing?

Ksivusya [100]

B) It is made
Of cells

4 0

3 years ago

Read 2 more answers

Which of these elements is unlikely to have a reaction with any element or compound?

Kay [80]

Argon is a noble gas. Argon has a full outer shell. This makes it so that it does not need to react with any of the other elements to be stable.

With Rubidium and Cobalt its a whole different story.

I hope that helps!

8 0

3 years ago

HELPPP

Ksivusya [100]

3 I’m pretty sure maybe

4 0

3 years ago

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0

4 years ago

Can someone solve the equations for these? 45 points

Volgvan

2)

formula equation: Pb(NO₃)₂(aq) + 2KI(aq) ----> PbI₂(s) + 2KNO₃(aq)
total ionic equation: Pb⁺² + 2NO₃⁻¹ + 2K⁺¹ + 2I⁻¹----> PbI₂(s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Pb⁺² + 2 I⁻¹----> PbI₂(s)

3)

formula equation: Zn(NO₃)₂(aq) + K₂CO₃(aq) -----> ZnCO₃(s) + 2KNO₃(aq)
total ionic equation: Zn⁺² + 2NO₃⁻¹ + 2K⁺¹ + CO₃⁻² ---> ZnCO₃ (s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Zn⁺² + CO₃⁻² ----> ZnCO₃ (s)

note: if I did not specify the state of the molecule in the reaction, you can assume they are aqueous unless state otherwise.

8 0

3 years ago