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4 years ago

The general format AB+CD→AD+BC represents a ____________ reaction.

Chemistry

1 answer:

geniusboy [140]4 years ago

3 0

The general format AB+CD→AD+BC represents are replacement reaction.

Explanation:

A replacement reaction transpires when elements rearrange positions in compounds. There are two kinds of replacement reactions: single and double.

A double replacement reaction happens when a pair of ionic compounds transactions. This creates a couple of new ionic compounds. A double replacement reaction can be described by the common equation:

AB + CD → AD + CB

AB and CD are the two reactant compounds, and AD and CB are the two product compounds that emerge from the reaction. Through the reaction, the ions B and D switch positions.

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What is the molarity of 2.3 mol of Kl dissolved in 0.5 L of water

Nastasia [14]

Answer:

$4.6\,\,moL\,\,L^{-1}$

Explanation:

Molarity refers to a measure of concentration.

Molarity = moles of solute/Litres of solution

Molarity refers to number of moles of solute present in this solution.

In order to find a solution's molarity, use value for the number of moles of solute and the total volume of the solution expressed in liters

As molarity of 2.3 mol of Kl is dissolved in 0.5 L of water,

Molarity = $\frac{2.3}{0.5} =4.6\,\,moL\,\,L^{-1}$

4 0

4 years ago

Methods: Part A: Preparation of Buffers Make two buffers starting with solid material, which is the most common way to make buff

Alecsey [184]

Answer:

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀ $\frac{A^{-} }{HA}$

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀ $\frac{HPO4^{2-} }{H2PO4^{-} }$

If desire pH is 7,0 you will obtain:

0,617 = $\frac{HPO4^{2-} }{H2PO4^{-} }$ (1)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)

Replacing (1) in (2) you will obtain:

[H₂PO₄⁻] = 0,0618 M

And with this value:

[HPO₄²⁻] = 0,0382 M

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× $\frac{0,0382mol}{1L}$ × $\frac{141,96g}{1mol}$ = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× $\frac{0,0618mol}{1L}$ × $\frac{119,96g}{1mol}$ = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀ $\frac{Tris-base }{Tris-H^{+} }$

If desire pH is 8,0 you will obtain:

0,841 = $\frac{Tris-base }{TrisH^{+} }$ (3)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] (4)

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× $\frac{0,0457mol}{1L}$ × $\frac{121,1g}{1mol}$ = 0,553 g of Tris-base

Tris-HCl = 0,1 L× $\frac{0,0543mol}{1L}$ × $\frac{157,6g}{1mol}$ = 0,856 g of Tris-HCl

I hope it helps!

8 0

3 years ago

Fundamental differences in the cell, the basic unit of life, allow for broadest classification of all living organisms into thre

Nana76 [90]

Answer:

Explanation:

7 0

3 years ago

How many electrons, protons and neutrons are in the element tin

lisabon 2012 [21]

Answer: Tin has 48 electrons, 69 neutrons, and 50 protons.

Explanation:

4 0

3 years ago

A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

Solving for V₂ , we get:

V₂ = 45.0 L

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0

4 years ago