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3 years ago

What is the electron configuration for potassium (K)?

Chemistry

1 answer:

Bond [772]3 years ago

3 0

Answer:

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:

Aufbau's principal of the arrangement of electronic configuration states that, electrons fill orbitals of lower energy first before the higher energy. 1s2 2s2 2p6 3s2 3p6 4s2 3d10...

This arrangement explains the Aufbau's principal, hence, or therefore, the 1s2 2s2 2p6 3s2 3p6 4s1 is the right answer for the representation of the electron configuration for potassium (K).

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Hydrogen is a possible future fuel. However, elemental hydrogen is rare, so it must be obtained from a hydrogen- containing comp

dimulka [17.4K]

Answer:

1.1 × 10² g

Explanation:

First, we will convert 1.0 L to cubic centimeters.

1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³

The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:

1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g

1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:

1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g

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3 years ago

1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n

Ede4ka [16]

Answer:

$m_{H2SO4$ = 61.25 g

$m_{Na2SO4}$ = 88.75 g

Explanation:

$m_{NaOH}$ = $\frac{200 . 25 }{100}$ = 50 g

⇒ $n_{NaOH}$ = $\frac{50}{40}$ = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

2 : 1 : 1 : 2

1.25 (moles)

⇒ $n_{H2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{H2SO4}$ = 0.625 × 98 = 61.25 g

$n_{Na2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{Na2SO4}$ = 0.625 × 142 = 88.75 g

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3 years ago

Consider the reversible reaction, A+B⇌C+D. If the concentration of product D is increased, the rate of the reverse reaction woul

natali 33 [55]

Answer:

If the concentration of product D is increased, the rate of the reverse reaction would increase.

Explanation:

Chemical reaction:

A + B ⇄ C + D

In given condition the equilibrium is disturb by increasing the concentration of product.

When the concentration of product D is increased the system will proceed in backward direction in order to regain the equilibrium. Because when the product concentration is high it means reaction is not on equilibrium state the reaction will proceed backward direction to regain the equilibrium state.

According to the Le- Chatelier principle,

At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.

The equilibrium can be disturb,

By changing the concentration

By changing the volume

By changing the pressure

By changing the temperature

4 0

3 years ago

Your teacher directs you to combine two liquid substances into a flask. after you combine them, you notice that the flask feels

adelina 88 [10]

Answer : The correct option is (A).

Explanation :

Endothermic reaction : When two liquids are combine into a flask, then the flask feels cold when we touch it because the system absorbed heat from the surrounding.

In general, endothermic process absorbs heat and cool the surrounding.

Exothermic reaction : When two liquids are combine into a flask, then the flask feels hot when we touch it because the system released heat into the surrounding.

In general, exothermic process releases heat and rise the temperature of surrounding.

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3 years ago

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$