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2 years ago

"1s22s22p63s2" is the electron configuration for which element

Chemistry

1 answer:

vazorg [7]2 years ago

7 0

MAGNESIUM(Mg) posses the up mentioned electronic configuration

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Select the correct answer. Given: 2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams

ivann1987 [24]

The reaction produces 2.93 g H₂.

M_r: 133.34 2.016

2Al + 6HCl → 2AlCl₃ + 3H₂

Moles of AlCl₃ = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

Moles of H₂ = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

Mass of H₂ = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

7 0

3 years ago

Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate

MatroZZZ [7]

Answer:

Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

3 0

3 years ago

Do metals gain or lose electrons as they form ions

Anna71 [15]

Answer:

They usually lose electrons to form ions with 2 positive charges

Explanation:

6 0

3 years ago

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A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

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2 years ago

What do you need to adjust to balance a chemical equation

Nataliya [291]

You always adjust numbers and it will be before the element
so you Neva add or change a subscript

8 0

3 years ago

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