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2 years ago

What is the solubility of MgCO₃ in a solution that contains 0.080 M Mg²⁺ ions? (Ksp of MgCO₃ is 3.5 × 10⁻⁸)

Chemistry

1 answer:

Alchen [17]2 years ago

7 0

Answer: 4.4 x 10^-7

Explanation:

The dissociation equation for this reaction is:

MgCO3 (s) → Mg+2 (aq) + CO3-2 (aq)

$\text { So, } k_{s p}=(x+0.08) x$ (Here 0.08 >>> x )

$\begin{aligned}\Rightarrow 3.5 \times 10^{-8} &=0.08 \times x \\\Rightarrow x &=\frac{3.5}{0.08} \times 10^{-8} \\&=4.37 \times 10^{-7} \\\Rightarrow x & \approx 4.4 \times 10^{-7} \mathrm{~M}\end{aligned}$

So the solubility MgCO₃ in a solution that containing 0.080 M Mg²⁺ is 4.4 x 10^-7

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‏A second - order reaction has a rate constant of 0.06 M min - 1 . It takes min for the reactant concentration to decrease from

alekssr [168]

Answer:

Choice D. Approximately $61.2$ minutes.

Explanation:

The reaction rate of a second-order reaction is proportional to the square of one of its reactants.

Let $y$ denote the concentration of that reactant (in $\rm M$ .)
Let $t$ denote time (in minutes.)

Let $k$ denote the rate constant of this reaction. Assume that $y_0$ is the concentration of that reactant at the beginning of this reaction (when $t = 0$ .) Because this reaction is of second order, it can be shown that:

$\displaystyle y = \frac{1}{k\, t + (1/y_0)}$ .

The question states that the rate constant here is $0.06\; \rm M\cdot min^{-1}$ . Hence, $k = 0.06$ . For simplicity, assume that $t = 0$ when the concentration is $0.13\; \rm M$ . Therefore, $y_0 = 0.13$ . The equation for concentration $y$ at time $t$ would become:

$\displaystyle y = \frac{1}{\underbrace{0.06}_{k}\, t + (1/\underbrace{0.13}_{y_0})}$ .

The goal is to find the time at which the concentration is $0.088$ . That's the same as solving this equation for $t$ , given that $y = 0.088$ :

$\displaystyle \frac{1}{0.06\; t + (1/0.13)} = 0.088$ .

$t \approx 61.2$ .

In other words, it would take approximately $61.2$ minutes before the concentration of this second-order reactant becomes $0.088\; \rm M$ .

6 0

3 years ago

1.Complete the balanced neutralization equation for the reaction below:

Alexxx [7]

1. $H_{2}$ S $O_{4}$ + Sr(OH) $_{2}$ ⇒ SrSO4 + 2 H2O is the balanced reaction.

2. 0.034 liters of KI will be required completely react with 2.43 g of Cu(NO3)2.

3. 0.55 liters is the volume in L of a 0.724 M Nal solution contains0.405 mol of NaI.

4. 33.3 ml many mL of 0.300 M NaF would be required to make a 0.0400 M solution of NaF when diluted to 250.0 mL with water

Explanation:

Balance chemical reaction of neutralization:

1. $H_{2}$ S $O_{4}$ + Sr(OH) $_{2}$ ⇒ SrSO4 + 2 H2O

2. Data given:

balance chemical reaction:

2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + I₂(s) + 4KNO₃(aq)

molarity of KI = 0.209 M

mass of Cu(NO₃)₂ = 2.43 grams

number of moles of Cu(NO₃)₂ will be calculated as:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

atomic mass of Cu(NO₃)₂ = 187.56 grams/mole

putting the values in the equation,

number of moles= $\frac{2.43}{187.56}$

= 0.0129 moles

2 moles of Cu(NO₃)₂ will react with 4 moles of KI

0.0129 moles will react with x moles of KI

$\frac{4}{2}$ = $\frac{x}{0.0129}$

x = 0.0258

atomic mass of KI = 166.00 grams/mole

mass = 166.00 x 0.0258

= 4.28 grams or ml is the final volume of KI

so molarity = $\frac{number of moles}{volume in litres}$

so molarity of KI is 0.0258 M, volume is 1 litre.

Using the formula

Minitial x Vinitial = M final Vfinal

V initial = $\frac{M final Vfinal}{Minitial}$

= $\frac{4.28 X 0.209}{0.0258}$

= 34.67 ml 0.034 liters of KI will be required.

3) Data given:

molarity of NaI = 0.724

number of moles of NaI =?

Volume in litres =?

formula used:

molarity = $\frac{number of moles}{volume in liters}$

volume in litres = $\frac{0.405}{0.724}$

= 0.55 liters is the volume

4) Data given:

Initial molarity = 0.3 M

initial volume = ?

final molarity = 0.04 M

final volume diluted by = 250 ml

formula used:

M initial X Vinitial = Mfinal X V final

putting the values in the equation:

Vinitial = $\frac{0.04 X 250}{0.3}$

= 33.3 ml of 0.3 M solution will be required.

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3 years ago

What is the [OH-]if the pH of a solution 10.62

Lerok [7]

As we know,

pH + pOH = 14
Or,
pOH = 14 - pH

Putting Value of pH,

pOH = 14 - 10.62

pOH = 3.38

pOH is converted into [OH⁻] as follow,

[OH⁻] = 10⁻pOH ∴ -pOH is in power, 10 means antilog
press shift and click log button
[OH⁻] = 10⁻³°³⁸

[OH⁻] = 0.0004168

[OH⁻] = 4.16 × 10⁻⁴

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3 years ago

A light wave has frequency of 4.5 * 10^19 Hz. How would this number appear on a scientific calculator? a. 4.E195 b. 19E4.5 c. 4.

scoray [572]

Answer:

Explanation:

3 0

3 years ago

Americium-241 is used in smoke detectors. it has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. by contras

Helga [31]

This question is missing the part that actually asks the question. The questions that are asked are as follows:

(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.

(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.

We can use the equation for a first order rate law to find the amount of material remaining after 4 days:

[A] = [A]₀e^(-kt)

[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.

(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.

4 days x 1 year/365 days = 0.0110

A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg

The decay of americium is so slow that no noticeable change occurs over 4 days.

(b) We can simply plug in the information of iodine-125 and solve for A:

A = (1.00)e^(-0.011 x 4)
A = 0.957 mg

Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.

7 0

3 years ago