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3 years ago

The 3 types of heat transfer are convection, radiation, connected. A. True B. False

Chemistry

1 answer:

frez [133]3 years ago

4 0

False. They are Convection, Conduction, and Radiation.

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1-The chemical potential energy of bond A is greater than the chemical potential energy of bond B. Which statement best explains

Tju [1.3M]

It should be noted that bond A has greater energy because C. The atoms in bond A are held more tightly together than the atoms in bond B.

The relationship between the bond energies of nitrogen, iodine, and fluorine gases is that the bond in nitrogen gas is the most difficult to break.

From the information given, the molecule with the greatest bid energy is CH4. The bind energy measures the bond strength that the chemical bond has.

Also, the bond energy of the reactants in reaction 1 is greater than the bond energy of the reactants in reaction 2. Due to this, reaction 1 requires a greater input of energy than reaction 2.

Lastly, the difference in the bond energy of Chlorine and Bromine is that Bromine has more electron levels than chlorine.

Learn more about bonds on:

brainly.com/question/819068

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3 years ago

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Identify what the conjugate acid/base would be. FOR H3PO4?

natali 33 [55]

Phosphoric acid. Also known as orthophosphoric acid in or phosphoric(V) acid, is a weak acid with the chemical formula H3PO4. The pure compound is a colorless solid.

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3 years ago

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .

Rewrite as a quadratic equation and solve for $x$ :

$x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)$

$x\approx 0.00111$ .

The pH of a solution depends on its H⁺ concentration.

At equilibrium

$[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}$ .

$\text{pH} = -\log{[\text{H}^{+}]} = 2.79$ .

5 0

3 years ago

ALSO NEED THIS ANSWERED PRETTY SOON I THINK IK WHAT IT IS BUT NOT SURE HELP!

Illusion [34]

Answer:

I think it's B but I could be wrong so really sorry if I am

3 0

2 years ago

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The steps required to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25x10^-2 M. Please

Yanka [14]

Answer:

In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.

Explanation:

Concentration: 1.25 x 10⁻² M

1,25 x 10⁻² mol FeCl₃ ___ 1000 mL

x ___ 200.0 mL

x = 2.5 x 10⁻³ mol FeCl₃

Mass of FeCl₃:

1 mol FeCl₃ _____________ 162.2 g

2.5 x 10⁻³ mol FeCl₃ _______ y

y = 0.4055 g FeCl₃

8 0

4 years ago