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3 years ago

What is 5.2034 x 10^8 in standard format

Chemistry

2 answers:

777dan777 [17]3 years ago

8 0

Five hundred twenty million, three hundred and forty thousand. hope it helps!

Elanso [62]3 years ago

6 0

The answer is 4162720
hop this works than anyone else

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If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the

Alina [70]

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

3 0

3 years ago

describe how knowledge of the periodic table would be important in three different careers, based on what you’ve read.

ANEK [815]

The knowledge of periodic table would be important in these three different careers such as medicine, pharmacology and engineering.

<h3>What is the importance of periodic table in following careers?</h3>

Periodic table is organised the whole elements vertically in groups and horizontally in periods with increasing atomic number.

The knowledge of arrangement of these elements serves alot of purposes in the following three different careers:

Medicine: The element, iron is an important metal in the periodic table used in treatment of low blood levels.

Pharmacology: The transition metals are useful in the field of pharmacology for the production of drugs. Example of such elements are chromium, manganese, cobalt, nickel, copper, and molybdenum.

Engineering: The information of elements in the periodic table helps engineers in designing of materials used for constructions.

Therefore, the knowledge of periodic table is very important in these three different careers such as medicine, pharmacology and engineering.

Learn more about period table here:

brainly.com/question/15987580

5 0

3 years ago

A cleaning bottle contains 83.1 g of ammonia. How many molecules of ammonia are in the bottle?

Tasya [4]

Answer:

2.94 x $10^2^4$

Explanation:

First we need to find out how many moles of ammonia there are, using the formula: Mass = mr x moles.

We know the mass is 83.1g, now we need to find the mR of ammonia - NH3.

N = 14, H = 1, so 14 + (3x1) = an mr of 17.

Moles = mass/ mr = 83.1/17 = 4.8882

Now we can multiply the moles by avogadro's constant to find the number of molecules:

4.8882 x (6.02 x $10^2^3$ ) = 2.94 x $10^2^4$ molecules of ammonia

4 0

3 years ago

How do waves form?

mrs_skeptik [129]

Answer:

It's the 4th one. waves are created by the wind

3 0

3 years ago

How do I do this? What are the answers to the 5 questions shown?

frozen [14]

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

6 0

3 years ago