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2 years ago

Which of the following does NOT demonstrate the law of conservation of matter?

Chemistry

1 answer:

Brrunno [24]2 years ago

8 0

The one that does not demonstrates the law of conservation of matter would be the third option: $2NO_2 + H_2O -- > HNO_3 + HNO_2$

<h3>What is law of conservation of matter? </h3>

It is a law that explains that matters are conserved during the course of chemical reactions. They can neither be destroyed nor created.

Thus, the number of moles of the species in a reaction remains the same before and after the reaction.

Thus, an equation that demonstrates the law of conservation of matter will be balanced in terms of the number of moles of species.

The only equation, in this case, is $2NO_2 + H_2O -- > HNO_3 + HNO_2$

There are 3 oxygen species on the reactant while 5 are present on the product side.

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1 year ago

Your classmate is viewing a sample using high power and is about to refocus using the coarse adjustment knob. What would you rec

Maurinko [17]

Answer:

Using the coarse adjustment knob of the microscope in high power may lead to the breaking of the slide if adjusted and raised the slide too much which can damage the sample as well as the high power lens.

In this case, I would recommend using the fine adjustment knob and moving away from the end of the viewing area of the microscope so there would no collision take place. The fine adjustment will help to get a clear image.

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2 years ago

Which organic compound is correctly matched with the subunit that composes it?

Alex787 [66]

It is the Starch-glucose. Glucose is a solitary sugar particle that your body can retain specifically in the digestive system. Sucrose and starches are starches shaped by at least two sugars reinforced together. The sugars in sucrose and starch must be separated into glucose particles in the gastrointestinal tract before your digestive organs can assimilate them.

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of

Olin [163]

Answer:

Empirical Formula = C₃H₆O₁

Solution:

Data Given:

Mass of Ethyl Butyrate = 3.61 mg = 0.00361 g

Mass of CO₂ = 8.22 mg = 0.00822 g

Mass of H₂O = 3.35 mg = 0.00335 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100

%C = (2.277) × (12 ÷ 44) × 100

%C = 2.277 × 0.2727 × 100

%C = 62.09 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100

%H = (0.9279) × (2.02 ÷ 18.02) × 100

%H = 0.9279 × 0.1120 × 100

%H = 10.39 %

%O = 100% - (%C + %H)

%O = 100% - (62.09% + 10.39%)

%O = 100% - 72.48%

%O = 27.52 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 62.09 ÷ 12.01

Moles of C = 5.169 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 10.39 ÷ 1.01

Moles of H = 10.287 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 27.52 ÷ 16.0

Moles of O = 1.720 mol

Step 3: Find out mole ratio and simplify it;

C H O

5.169 10.287 1.720

5.169/1.720 10.287/1.720 1.720/1.720

3.00 5.98 1

3 ≈ 6 1

Result:

Empirical Formula = C₃H₆O₁

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3 years ago

How does the chemical formula for water influence its chemical properties?

OLga [1]

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