All categories

3 years ago

Which ion would represent the ion of an element from Group 2A?

Chemistry

1 answer:

ziro4ka [17]3 years ago

3 0

Answer:

E²⁺

Explanation:

The group two contain alkaline earth metals.

There are six elements in group 2A.

Beryllium, Magnesium, calcium, strontium, barium and radium.

All members have two valance electrons.

They lose two valance electrons to complete the octet.

When they lose the two valance electrons they form cation X²⁺.

They react with halogens and form salt such as

MgCl₂, CaCl₂ etc.

Mg²⁺ Cl²⁻₂

The oxidation state of halogens are -1, while the elements of group two A shows +2 that's why two atoms of halogen are combine with one atom of alkaline earth metals and make the compound overall neutral.

All the alkaline earth metals have similar properties.

You might be interested in

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached

Rashid [163]

Answer:

(a) Cu²⁺ +2e⁻ ⇌ Cu

Explanation:

(a) Cu half-reaction

Cu²⁺ + 2e⁻ ⇌ Cu

(c) Cell voltage

The standard reduction potentials for the half-reactions are+

E°/V

Cu²⁺ + 2e⁻ ⇌ Cu; 0.34

Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241

The equation for the cell reaction is

E°/V

Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34

2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; -0.241

Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10

The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation

(ii) Calculations:

T = 25 + 273.15 = 298.15 K

$Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}$

3 0

3 years ago

0 ml from a 2.0 m solution of hcl is diluted to 10.0 ml using deionized water. what is the concentration of the new solution?

krok68 [10]

The concentration would remain at 2.0m. The problem states that 0 ml from the 2.0 m solution is diluted, therefore implying that none of it was diluted. Therefore, the level of concentration of the new solution would be the same as before.

3 0

3 years ago

Which is a characteristic of mixtures ?

sattari [20]

The characteristics of mixtures are:

a) they are made up of two of more substances

b) the two substances can be separated by physical means

c) there are two kinds of mixtures : homogenous and heterogenous

d) homogenous mixtures do not have any physicaly distinguished boundary

6 0

4 years ago

Read 2 more answers

Let us assume that fe(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented

Yuki888 [10]

17.8 mL NaOH

Step 1. Write the chemical equation

Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]

= 11.50 mmol Fe^(2+)

Step 3. Calculate the moles of NaOH

Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]

= 23.00 mmol NaOH

Step 4. Calculate the volume of NaOH

Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)

= 17.8 mL NaOH

3 0

4 years ago

Student A performed gravimetric analysis for sulfate in her unknown using the same procedures we did . Results of her three tria

Andre45 [30]

Let us first calculate the accepted value of sulfate using the molar mass of sodium sulfate.

The molar mass of sodium sulfate is 2 Na + S + 4 O = 2 (22.99) + 32.07 + 4 (16)

Molar mass of Sodium sulfate = 142.05

Molar mass of sulfate = S + 4 (O) = 32.07 + 4 (16) = 96.07

% of sulfate in sodium sulfate = (Molar mass of SO4 / Molar mass of Na2SO4) x 100

% of sulfate = (96.07/142.05) x 100

% of sulfate = 67.6%

Theoretical value for % Sulfate is 67.6%

Percent error is calculated as

Percent Error = $\left | \frac{Experimental - Theoretical}{Theoretical} \right |$

Let us calculate mean for student A.

Mean for student A = (68.6 + 66.2 + 67.1) / 3 = 201.9/3 = 67.3 %

% error for student A = $\left | \frac{67.3 - 67.6}{67.6} \right |$

% error for student A = 0.44%

Mean for student B = (66.7 +66.6 + 66.5)/3 = 199.8 /3 = 66.6 %

% error for student B = $\left | \frac{66.6 - 67.6}{67.6} \right |$

% error for student B = 1.48%

Accuracy is the closeness of experimental values to the true value. This is defined in terms of absolute and percent errors.

Since percent error for student A is lower, we can say that student A was more accurate.

The precision of the data depends on the closeness of experimental values to each other. We can see that experimental values for student B were close to each other.

Therefore we can say that student B was more precise.

8 0

3 years ago