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2 years ago

What occurs when potassium reacts with chlorine to form potassium chloride?

Chemistry

2 answers:

djverab [1.8K]2 years ago

7 0

Answer: The electrons get transferred from potassium to chlorine to form potassium chloride

Explanation:

Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.

The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.

Potassium is the 19th element of the periodic table having electronic configuration of $1s^22s^22p^63s^23p^64s^1$

This element will loose 1 electron to form $K^+$ ion

Chlorine is the 17th element of the periodic table having electronic configuration of $1s^22s^22p^63s^23p^5$

This element will gain 1 electron to form $Cl^-$ ion

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

The ionic compound formed will have a chemical formula of KCl

Hence, the electrons get transferred from potassium to chlorine to form potassium chloride

schepotkina [342]2 years ago

6 0

The potassium will donate one of its valence electrons

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The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p

saul85 [17]

Answer : The vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of bromine at $10.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $10.0^oC=273+10.0=283.0K$

$T_2$ = normal boiling point of bromine = $58.8^oC=273+58.8=331.8K$

$\Delta H_{vap}$ = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})$

$P_1=0.1448atm$

Hence, the vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

4 0

3 years ago

how many calories is in one peanut if the volume of water is 10 mL and the water temperature is rise 4 degrees celsius

liraira [26]

Answer:

40.02 calories

Explanation:

V = 10 mL = 10g

we know t went up by 4°C, this is our ∆t as it is a change.

Formula that ties it together: Q = mc∆t

where,

Q = energy absorbed by water

m = mass of water

c = specific heat of water (constant)

∆t = temperature change

Q = (10 g) x (4.186 J/g•°C) x (4°C)

Q = 167.44 J

Joules to Calories:

167.44 J x 1 cal/4.184 J = 40.02 calories

(makes sense as in image it is close to the value).

4 0

3 years ago

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

pav-90 [236]

Answer: The initial concentration of hydrogen peroxide at the given temperature is 0.399 M

Explanation:

Decomposition of hydrogen peroxide is following first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $2.54\times 10^{-4}s^{-1}$

t = time taken for decay process = 855 s

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.321 M

Putting values in above equation, we get:

$2.54\times 10^{-4}s^{-1}=\frac{2.303}{855s}\log \frac{[A_o]}{0.321}$

$[A_o]=0.399M$

Hence, the initial concentration of hydrogen peroxide at the given temperature is 0.399 M

7 0

3 years ago

HELPPPPPPP is this chemical or physical reaction

deff fn [24]

Answer:

Chemical reaction

Explanation:

It's creating a whole other substance

7 0

2 years ago

Caffeine, a stimulant found in coffee and soda, has the following elemental composition: 49.48% carbon, 5.19% hydrogen, 16.48% o

prisoha [69]

Answer:

Explanation:

C = 49.48

H = 5.19

O = 16.48

N = 28.85

ratio of moles

= 49.48 / 12 : 5.19 / 1 : 16.48 / 16 : 28.85 / 14

= 4.123 : 5.19 : 1.03 : 2.06

= 4 : 5 : 1 : 2

so the empirical formula = C₄ H₅O N₂

Let molecular formula = ( C₄ H₅ON₂ )ₙ ,

n ( 48 + 5 + 16 + 28 ) = 119.19

97 n = 194.19

n = 2 ( approx )

molecular formula = C₈ H₁₀O₂ N₄

3 0

3 years ago