When the following redox equation is balanced with smallest whole number coefficients, the coefficient for Sn(OH)3– will be _____. Bi(OH)3(s) + Sn(OH)3–(aq) → Sn(OH)62–(aq) + Bi(s) (basic solution)

Answers

Answer 1

Answer:

Answer:

The coefficient of $Sn(OH)_3^(-)$ is 3 in the balanced redox reaction.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom loses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^(n+)+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^(n+)+ne^-\rightarrow X$

For the given chemical reaction:

$Bi(OH)_3+Sn(OH)_3^(-)\rightarrow Sn(OH)6^(2-)+Bi$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $Sn(OH)_3^(-)+3OH^-\rightarrow Sn(OH)6^(2-)+2e^-$

Reduction half reaction: $Bi(OH)_3+3e^-\rightarrow Bi+3OH^-$

To balance the oxidation half reaction must be multiplied by 3 and reduction half reaction must be multiplied by 2 thus, the balanced equation is:-

$3Sn(OH)_3^(-)+3OH^-+2Bi(OH)_3\rightarrow 3Sn(OH)6^(2-)+2Bi$

The coefficient of $Sn(OH)_3^(-)$ is 3 in the balanced redox reaction.

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Determine the (H+), pH, and pOH of a solution with an [OH-] of 9.5 x 10-10 M at 25 °C. M pH =

Answers

Answer : The concentration of $H^+$ ion, pH and pOH of solution is, $1.05* 10^(-5)M$ , 4.98 and 9.02 respectively.

Explanation : Given,

Concentration of $OH^-$ ion = $9.5* 10^(-10)M$

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pOH=-\log [OH^-]$

$pOH=-\log (9.5* 10^(-10))$

$pOH=9.02$

The pOH of the solution is, 9.02

Now we have to calculate the pH.

$pH+pOH=14\n\npH=14-pOH\n\npH=14-9.02=4.98$

The pH of the solution is, 4.98

Now we have to calculate the $H^+$ concentration.

$pH=-\log [H^+]$

$4.98=-\log [H^+]$

$[H^+]=1.05* 10^(-5)M$

The $H^+$ concentration is, $1.05* 10^(-5)M$

Answer:

pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978

Explanation:

Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C

As, pOH = - log [OH⁻]

⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022

The self-ionisation constant of water is given by

Kw = [H⁺] [OH⁻] and pKw = pH + pOH

Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.

Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴

⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M

also,

pH + pOH = pKw = 14

⇒ pH = 14 - pOH = 14 - 9.022 = 4.978

Iodine-131, t1/2 = 8.0 days, is used in diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-131 initially emits 9.95 × 1018 β particles per day, how long will it take for the activity to drop to 6.22 × 1017 β particles per day?

Answers

Explanation:

Formula for the first order decay is as follows.

$ln((A)/(A_(o)))$ = -kt

where, A = activity at time t

$A_(o)$ = initial activity

k = decay constant

Hence, putting the given values into the above formula as follows.

k = $\frac{ln(2)}{\text{half life}}$

= $(ln(2))/(8.0)$

= 0.086643 per day

Also, $(ln(6.22 * 10^(17)))/(9.95 * 10^(8)) = -0.086643 * t$

t = 32 days

Thus, we can conclude that it will take 32 days for the activity to drop to $6.22 * 10^(17)$ $\beta$ particles per day.

Consider the reaction of aqueous potassium sulfate with aqueous g silver nitrate based on the solubility rule predict the product likely to be precipitate write a balanced molecular equation describing the reaction.

Answers

Answer:

K₂SO₄(aq) + 2AgNO₃ (aq) → 2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

Explanation:

Our reactants are: K₂SO₄ and AgNO₃

By the solubility rules, we know that sulfates are insoluble when they react to Ag⁺, Pb²⁺, Ca²⁺, Ba²⁺, Sr²⁺, Hg⁺

We also determine, that salts from nitrate are all soluble.

The reaction is:

K₂SO₄(aq) + 2AgNO₃ (aq) → 2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

In metallic bonds, the mobile electrons surrounding the positive ions are called a(n)

Answers

It is called, Noble gas

A food is initially at a moisture content of 90% dry basis. Calculate the moisture content in wet basis

Answers

Answer:

Moisture content in wet basis = 47.4 %

Explanation:

Moisture content expresses the amount of water present in a moist sample.Dry basis and wet basis are widely used to express moisture content.

The next equation express the moisture content in wet basis:

$MC_(wb)=(MC_(db))/(1+MC_(db))$

where, $MC_(wb)$ : moisture content in wet basis and

$MC_(db)$ : moisture content in dry basis

We now calculate the moisture content in wet basis:

$MC_(wb)=(MC_(db))/(1+MC_(db))$

$MC_(wb)=(0.90)/(1+0.90)$

$MC_(wb)$ = 0.474 = 47.4 % wet basis

Have a nice day!

If the crucible originally weighs 3.715 g and 2. 000 g of hydrate are added to it , what is the weight of the water that is lost if the final weight of the crucible and anhydrous salt is 5.022?

Answers

For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.

From the problem, you know that 1) = 2.000 g.

Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).

1) = 5.022 g - 3.715 g = 1.307 g

Now you can complete our original task.

Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.

Final answer:

The weight of the water lost is 0.693 g.

Explanation:

To calculate the weight of the water that is lost, we need to find the weight of the anhydrous salt. The anhydrous salt is the crucible with the added hydrate, minus the weight of the crucible. So, the weight of the anhydrous salt is 5.022 g - 3.715 g = 1.307 g. Since the weight of the hydrate is 2.000 g, the weight of the water that is lost is equal to the difference between the weight of the hydrate and the weight of the anhydrous salt, which is 2.000 g - 1.307 g = 0.693 g.

Learn more about Hydrates here:

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Please help....thank you