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xxTIMURxx [149]
2 years ago
6

Someone who occupies a key role in the design, construction, and management of chemical operations to insure and improve the qua

lity of life for people in and outside the plant is a(n) _____.
A chemical technician
B information chemist
C safety engineer
D environmental chemist
Chemistry
1 answer:
patriot [66]2 years ago
8 0

Answer:

D: environmental chemsit

Explanation:

I just read an article for my chemistry class, and took the quiz... hope this helps:)

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How many grams of HNO3 are produced when 59.0 g of NO2 completely reacts?
Anettt [7]

Answer:

53.7 grams of HNO3 will be produced

Explanation:

Step 1: Data given

Mass of NO2 = 59.0 grams

Molar mass NO2 = 46.0 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 59.0 grams / 46.0 g/mol

Moles NO2 = 1.28 moles

Step 4: Calculate moles HNO3

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3

Step 7: Calculate mass HNO3

Mass HNO3 = 0.853 moles * 63.01 g/mol

Mass HNO3 = 53.7 grams

53.7 grams of HNO3 will be produced

3 0
3 years ago
A student was asked to find the molar mass of calcium nitrate, Ca(NO3)2. she added the molar masses of calcium, nitrogen, and ox
dlinn [17]

Answer: 116g/mole

Explanation:

She didn't get the answer because she didn't add the them well , due to the bracket present.

3 0
2 years ago
How are conclusions and evidence related
serg [7]
You need evidence to support a conclustion 
3 0
2 years ago
Read 2 more answers
What advice did Planck ignore when he was a student,and how did it impact future discoveries
yan [13]

Answer:

A PROFESSOR WARNED HIM NOT TO GO INTO PHYSICS. Not long after the 16-year-old Planck got to the University of Munich in 1874, physics professor Philipp von Jolly tried to dissuade the young student from going into theoretical physics. Jolly argued that other scientists had basically figured out all there was to know.

Explanation:

5 0
3 years ago
Determine the empirical formulas for compounds with the following percent compositions:
Elis [28]

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is P_2O_5

<u>For b:</u> The empirical formula of the compound is KH_2PO_4

<u>Explanation:</u>

  • <u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = \frac{1.406}{1.406}=1

For Oxygen = \frac{3.525}{1.406}=2.5

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = 1\times 2=2

For Oxygen = 2.5\times 2=5

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is P_2O_5

  • <u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium =\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = \frac{0.736}{0.735}=1

For Hydrogen = \frac{1.5}{0.735}=2.04\approx 2

For Phosphorus = \frac{0.735}{0.735}=1

For Oxygen = \frac{2.9375}{0.735}=3.99\approx 4

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is KH_2PO_4

3 0
3 years ago
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