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serious [3.7K]
2 years ago
6

CaCl2 (s) + 2 H2O → Ca(OH)2 (aq) + 2 HCl (g) When Addie added CaCl2 to the water in her flask, she noted that the flask heated u

p. What conclusion can be drawn about the bond energy of products and reactants? A) More energy was absorbed in the formation of chemical bonds than released in the breaking of chemical bonds. B) Less energy was absorbed in the breaking of chemical bonds than released in the formation of chemical bonds. C) More energy was absorbed in the breaking of chemical bonds than released in the formation of chemical bonds. D) Less energy was absorbed in the formation of chemical bonds than released in the breaking of chemical bonds.
Chemistry
1 answer:
zloy xaker [14]2 years ago
3 0

Answer: the answer is b

Explanation: the formation of bonds releases energy and the breaking of bonds always involves an absorption of energy

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Be sure to answer all parts.
Liula [17]

A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²

B. The cost of the patch to the nearest cent is 734 cents

<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>

We can convert 16.1 cm² to km² as illustrated below:

Conversion scale

1 cm² = 1×10⁻¹⁰ km²

Therefore,

16.1 cm² = 16.1 × 1×10⁻¹⁰

16.1 cm² = 1.61×10⁻⁹ km²

Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²

<h3>B. How to determine the cost in cent</h3>

We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:

1 cm² = 0.155 in²

Therefore,

16.1 cm² = 16.1 × 0.155

16.1 cm² = 2.4955 in²

Finally, we shall the determine the cost in centas fo r llow:

  • Cost per in² = $2.94 = 294 cent
  • Cost of 2.4955 in² =?

1 in² = 294 cent

Therefore,

2.4955 in² = 2.4955 × 294

2.4955 in² = 734 cents

Thus, the cost of the patch is 734 cents

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

6 0
1 year ago
Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3
Vadim26 [7]
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
7 0
3 years ago
Read 2 more answers
PLEASE HELP!
Anon25 [30]

Answer:

C. Mole

Explanation:

Chemical formula deals with mole number

4 0
2 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
2 years ago
How can you locate a family of elements in the table? Name all the elements in the oxygen family.
ANEK [815]
<span>every column on the table represents a family that react similarly with other <span>elements.</span></span>
4 0
2 years ago
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