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3 years ago

If x represents the molar solubility of b a 3 ( p o 4 ) 2 , what is the correct equation for the k s p ?

Mathematics

1 answer:

AnnyKZ [126]3 years ago

5 0

The molar solubility of Ba₃(PO₄)₂ is its concentration of pure substance

The correct equation for the ksp is Ksp = 6x⁵

<h3>How to determine the equation of ksp?</h3>

Start by writing the reaction for the solution of Barium phosphate

Ba₃(PO₄)₂ ⇌3Ba²⁺ +2PO₄³⁻

Next, we make an ICE chart

Let "x" be the molar solubility of Barium phosphate

Ba₃(PO₄)₂(s) ⇌3Ba²⁺ +2PO₄³⁻

I 0 0

C +3x +2x

E 3x³ 2x²

Lastly, we write the expression for the solubility product constant (Ksp)

This is represented as:

Ksp = [3Ba²⁺] × [2PO₄³⁻]²

Ksp = 3x³ × 2x²

Evaluate the product

Ksp = 6x⁵

Hence, the correct equation for the ksp is Ksp = 6x⁵

Read more about molar solubility at:

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Area of the bounded curves y=x^2, y=√(7+x)

N76 [4]

Answer:

$\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 5.74773$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

$\displaystyle \left \{ {{y = x^2} \atop {y = \sqrt{7 + x}}} \right.$

Step 2: Identify

Graph the systems of equations - see attachment.

Top Function: $\displaystyle y = \sqrt{7 + x}$

Bottom Function: $\displaystyle y = x^2$

Bounds of Integration: [-1.529, 1.718]

Step 3: Integrate Pt. 1

Substitute in variables [Area of a Region Formula]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - \int\limits^{1.718}_{-1.529} {x^2} \, dx$
[Right Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - \frac{x^3}{3} \bigg| \limits^{1.718}_{-1.529}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - 2.88176$

Step 4: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 7 + x$
[u] Basic Power Rule [Derivative Rule - Addition/Subtraction]: $\displaystyle du = dx$
[Limits] Switch: $\displaystyle \left \{ {{x = 1.718 ,\ u = 7 + 1.718 = 8.718} \atop {x = -1.529 ,\ u = 7 - 1.529 = 5.471}} \right.$

Step 5: Integrate Pt. 3

[Integral] U-Substitution: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{8.718}_{5.471} {\sqrt{u}} \, du - 2.88176$
[Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = \frac{2x^\Big{\frac{3}{2}}}{3} \bigg| \limits^{8.718}_{5.471} - 2.88176$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 8.62949 - 2.88176$
Simplify: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 5.74773$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

5 0

3 years ago