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iragen [17]
3 years ago
12

Which element would lose electrons during ionic bonding choose all that apply

Chemistry
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

All of them!

Explanation:

Since Mg, Li, Ca, and Cs are all in groups 1 and 2 of the periodic table, they are alkali/alkaline earth metals and will all lose electrons during ionic bonding.

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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t
ratelena [41]

Answer:

\Delta G^0 _{rxn} = 207.6\ kJ/mol

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

\Delta G_f^0  \ \ \ N_2O_{(g)} = 103 .8  \ kJ/mol

\Delta G_f^0  \ \ \ N_2{(g)} =0 \ kJ/mol

\Delta G_f^0  \ \ \ O_2{(g)} =0 \ kJ/mol

\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times  0 +   0)

\Delta G^0 _{rxn} = 207.6\ kJ/mol

The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}

K_p = \dfrac{1}{ (22.20)}

K_p = 0.045

\Delta G = \Delta G^0 _{rxn} + RT \ lnK

where;

R = gas constant = 8.314 × 10⁻³ kJ

\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

ΔG ≅ 199.91 kJ

7 0
4 years ago
Write the reaction for the weak base co32- in water
Gekata [30.6K]

The weak base carbonate reacts with the water forming bicarbonate and the hydroxide ion.

<h3>What is a weak base?</h3>

The weak bases are given as the compounds that have the inability to get completely dissociated in water.

The carbonate is given as the weak base, as in water it remains constituted in the form of a bond with oxygen.

The reaction of water with the carbonate ion is given as:

\rm CO_3^{2-}\;+\;H_2O\;\rightleftharpoons HCO_3^-\;+\;OH^-

Learn more about base, here:

brainly.com/question/23687757

#SPJ4

3 0
2 years ago
PLEASE HELP URGENT!!! ( photo included)
enyata [817]

Answer:

1. open

2. open

Explanation:

1. An open system can exchange energy and matter.

2. The ocean is known to be an open system.

8 0
3 years ago
Determine the best course of action when acid or base is spilled on the skin. Apply base to an acid burn, or acid to a base burn
Hunter-Best [27]

Answer:

You have to apply the base to an acid burn and vice versa.

Explanation:

4 0
3 years ago
Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.
natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

pH=4.76+log(\frac{0.95}{0.92})

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

pH=4.76+log(\frac{0.99}{0.88})

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

pH=4.76+log(\frac{0.85}{1.02})

pH = 4.76 - 0.079

pH = 4.681

5 0
4 years ago
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