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3 years ago

What principle requires that a chemical equation be balanced?

1 answer:

7 0

The principle that requires that a chemical equation be balanced would be the law of definite proportions. It <span>states that a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation. Hope this answers the question.</span>

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Dafna11 [192]

Answer:

c and d are correct

Explanation:

In A, false because in Valence Electrons, the more the valences, the more stable an atom is.

In B, false because atoms cannot readily gain or lose valence electrons as the number of valence electrons is determined by the column they are in.

In C, true because the more the valence electrons, the more the stability of an atom.

In D, true as electron placing is important and the reactivity of an atom is important.

So C and D are true!

7 0

2 years ago

Read 2 more answers

What would be the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg?

Anna [14]

Answer: The temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 2300 mm Hg = 3.02 atm (760mmHg=1atm)

V = Volume of gas = 15 L

n = number of moles = 0.6

R = gas constant = $0.0821Latm/Kmol$

T =temperature = ?

$T=\frac{PV}{nR}$

$T=\frac{3.02atm\times 15L}{0.0821Latm/K mol\times 0.6mol}=920K$

Thus the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K

8 0

3 years ago

What wave has the longer wavelength

natima [27]

The wavelength is the distance between one crest/trough to another crest/trough. On the image, it's basically the length between each peak of the wave. You can see that the distance between peaks in wave A are much shorter than the distance between the peaks in wave B.

Thus, wave B has the longer wavelength.

8 0

2 years ago

How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution

dalvyx [7]

<h3>Answer:</h3>

0.024 kg CaO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Aqueous Solutions</u>

Molarity = moles of solute / liters of solution

<u>Atomic Structure</u>

Reading a Periodic Tables
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

<u>Step 2: Identify Conversions</u>

1000 g = 1 kg

Molar Mass of Ca - 40.08 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})$
Multiply: $\displaystyle 0.024114 \ kg \ CaO$