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3 years ago

9

, you will change the temperature and particle size of the antacid tablet and observe how these changes affect the reaction. In

the space below, write a scientific question that you will answer by doing this experiment.

1 answer:

snow_lady [41]3 years ago

8 0

Answer:

i dont remember

Explanation:

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A client takes 30 ml of magnesium hydroxide and aluminum hydroxide with simethicone P.O. 1 hour and 3 hours after each meal and

labwork [276]

Answer:

Due to the short term of its action it has in the stomach environment

Explanation:

Aluminum hydroxide, magnesium hydroxide, and simethicone is a combination drug used for the treatment of upset stomach, acid indigestion, bloating heartburn caused by gas, or stomach discomfort caused by eating or drinking too much

4 0

3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s) + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s) + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s) + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0

3 years ago

How many grams of Ca(NO3)2 are needed to make 25.0 g of a 15.0% Ca(NO3)2(aq)?

yarga [219]

Answer:

3.75 g.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em />

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of calcium nitrite = ??? g,

mass of the solution = 25.0 g.

∴ mass % = (mass of solute/mass of solution) x 100

<em></em>

<em>∴ mass of solute (calcium nitrite) = (mass %)(mass of solution)/100</em> = (15.0 %)(25.0 g)/100 = <em>3.75 g.</em>

5 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.18 M solution of H₂A?

Tcecarenko [31]

Answer:

pH = 2.10

Explanation:

We name an acid as diprotic because it can release two protons:

H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁

HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂

We propose the mass balance:

Analytical concentration = [H₂A] + [HA⁻] + [A⁻²]

As Ka₂ is so small, we avoid the [A⁻²] so:

0.18 M = [H₂A] + [HA⁻]

But we can not avoid the HA⁻, because the Ka₁. Ka₁'s expression is:

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

We propose the charge balance:

[H₃O⁺] = [HA⁻] + [A⁻²] + [OH⁻]

As we did not consider the A⁻², we can miss the term and if

Kw = H⁺ . OH⁻

We replace Kw/H⁺ = OH⁻. So the new equation is:

[H₃O⁺] = [HA⁻] + Kw / [H₃O⁺]

The acid is so concentrated, so we can avoid the term with the Kw, so:

[H₃O⁺] = [HA⁻]

In the mass balance we would have:

0.18 M = [H₂A]

We replace at Ka₁

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

Ka1 . 0.18 / [H₃O⁺] = [HA⁻]

We replace at the charge balance:

[H₃O⁺] = Ka1 . 0.18 / [H₃O⁺]

[H₃O⁺]² = 3.4×10⁻⁴ . 0.18

[H₃O⁺] = √(3.4×10⁻⁴ . 0.18)

[H₃O⁺] = 7.82×10⁻³

- log [H₃O⁺] = pH → - log 7.82×10⁻³

pH = 2.10

5 0

3 years ago