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3 years ago

Which element's properties are similar to potassium?

Chemistry

2 answers:

12345 [234]3 years ago

6 0

Answer:

Potassium is an alkali metal

Explanation:

Symbol: K

Atomic number: 19

Atomic mass: 39.0983 u

Electron configuration: [Ar] 4s¹

Van der Waals radius: 280 pm

Electronegativity: 0.82

vekshin13 years ago

5 0

Answer:

Lithium, Sodium, Rubidium, Caesium, and Francium.

Explanation:

Like Potassium (K), those elements are alkali metals, found in Group 1 of the periodic table. Elements of a group share similar chemical properties with each other. So, any of those should work.

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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

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3 years ago

A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

Step 1: Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

Step 2: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

Step 3: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

Step 4: When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

Step 5: Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

Step 6: Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

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Andrej [43]

Answer:

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Explanation:

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4 years ago