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labwork [276]
2 years ago
8

The net ionic equation for formation of an aqueous solution of NiI 2 accompanied by evolution of CO 2 gas via mixing solid NiCO

3 and aqueous hydriodic acid is ________. 2NiCO3 (s) HI (aq) 2H2O (l) CO2 (g) 2Ni2 (aq) NiCO3 (s) I- (aq) 2H2O (l) CO2 (g) Ni2 (aq) HI (aq) NiCO3 (s) 2H (aq) H2O (l) CO2 (g) Ni2 (aq) NiCO3 (s) 2HI (aq) 2H2O (l) CO2 (g) NiI2 (aq) NiCO3 (s) 2HI (aq) H2O (l) CO2 (g) Ni2 (aq) 2I- (aq)
Chemistry
1 answer:
Naily [24]2 years ago
6 0

Answer:NiCO_3(s)+2H^+(aq)\rightarrow Ni^{2+}(aq)+CO_2(g)+H_2O(l)

Explanation:

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The balanced chemical equation will be:

NiCO_3(s)+2HI(aq)\rightarrow NiI_2(aq)+CO_2(g)+H_2O(l)

The total ionic chemical equation will be:

NiCO_3(s)+2H^+(aq)+2I^-(aq)\rightarrow Ni^{2+}(aq)+2I^-(aq)+CO_2(g)+H_2O(l)

The ions which are present on both the sides of the equation are iodide ions and hence are not involved in net ionic equation.

NiCO_3(s)+2H^+(aq)\rightarrow Ni^{2+}(aq)+CO_2(g)+H_2O(l)

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What is the pH of a solution made by mixing 15.00 mL of 0.100 M HCl with 50.00 mL of 0.100 M KOH? Assume that the volumes of the
denis23 [38]

Answer:

The correct answer is: pH = 12.73

Explanation:

The <em>neutralization reaction</em> between HCl and KOH is given by the following chemical equation:

HCl + KOH ⇒ KCl + H₂O

Since HCl is a strong acid and KOH is a strong base, HCl is completely dissociated into H⁺ and Cl⁻ ions, whereas KOH is dissociated completely into K⁺ and OH⁻ ions.

For acids, the number of equivalents is given by the moles of H⁺ ions (in this case: 1 equivalent per mol of HCl). For bases, the number of equivalents is given by the moles of OH⁻ ions (in this case: 1 equivalent per mol of KOH).

The H⁺ ions from HCl will react with OH⁻ ions of KOH to give H₂O. The pH is calculated from the difference between the equivalents of H⁺ and OH⁻:

equivalents of H⁺= volume HCl x Molarity HCl

                            = (15.0 mL x 1 L/1000 mL) x 0.100 mol/L

                            = 1.5 x 10⁻³ eq H⁺

equivalents of OH⁻= volume KOH x Molarity KOH

                               = (50.0 mL x 1 L/1000 mL) X 0.100 mol/L

                               = 5 x 10⁻³ eq OH⁻

There are more OH⁻ ions than H⁺ ions. The excess of OH⁻ (that did not react with H⁺ ions) is calculated as follows:

OH⁻ ions= (5 x 10⁻³ eq OH⁻) -  (1.5 x 10⁻³ eq H⁺) = 3.5 x 10⁻³ eq OH⁻= 3.5 x 10⁻³ moles OH⁻  

As the volumes of the solutions are additive, the total volume of the solution is:

V= 15.0 mL + 50.0 mL = 65.0 mL= 0.065 L

So, the concentration of OH⁻ ions in the solution is given by:

[OH⁻] = moles OH⁻/V= (3.5 x 10⁻³ moles OH⁻)/0.065 L = 0.054 mol/L = 0.054 M  

From  [OH⁻], we can calculate pOH:

pOH = -log [OH⁻] = -log (0.054) = 1.27

Finally, we know that pH + pOH= 14; so we calculate pH:

pH= 14 - pOH = 14 - 1,27 =  12.73                                                            

8 0
3 years ago
The percent by mass of copper in CuBr2 is Use mc004-1.jpg. 28.45%. 44.30%. 63.55%. 71.55%.
Greeley [361]
First, we must find the total mass of CuBr₂:
Mass = 64 + 2 x 80
Mass = 224

Percentage mass of copper = mass of copper x 100 / total mass
Percentage mass of copper = (64 / 224) x 100
Percentage mass of copper = 28.45%
3 0
2 years ago
Read 2 more answers
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
How many moles are there in 45.0 grams of sulfuric acid,<br> H2SO4?
olga2289 [7]

Answer:

the answer is 0.4588162459

Explanation:

1 mole = 0.010195916576195

= 45 ×"

5 0
2 years ago
In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OAT
dmitriy555 [2]

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy  

\Delta H = enthalpy change

\Delta S = entropy change  

T = temperature in Kelvin

When \Delta G = +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

For the given reaction 1: glucose+Pi\rightarrow glucose-6-phosphate+H_2O \Delta G=+13.8kJ/mol

As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

6 0
2 years ago
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