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Likurg_2 [28]
3 years ago
15

Which statement is correct?

Chemistry
1 answer:
rosijanka [135]3 years ago
4 0

Answer:

B. Chemical changes produce new substances; physical changes do not.

Explanation:

Physical changes only alter appearances, while chemical changes are interactions that create new chemical formulas.

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Which describes the role of oxygen in photosynthesis and cellular respiration?
Marianna [84]

photosynthesis makes the glucose that is used in cellular respiration to make ATP. The glucose is then turned back into carbon dioxide, which is used in photosynthesis. While water is broken down to form oxygen during photosynthesis, in cellular respiration oxygen is combined with hydrogen to form water.

4 0
4 years ago
Analytical chemistry has two branches, _____, that are important to forensic chemistry. qualitative and quantitative qualitative
tatuchka [14]

Answer:

qualitative and quantitative

Explanation:

8 0
2 years ago
Which substance can be broken down by chemical means
Maru [420]

Answer:

1. Ammonia  

Step-by-step explanation:

Ammonia, NH₃, is a <em>compound</em> of nitrogen and hydrogen.

Thus, it can be broken down by chemical means into the <em>elements</em> nitrogen and hydrogen.

The other options are <em>wrong</em>, because Al, Sb, and Pb are elements. They cannot be broken down chemically into simpler substances.

4 0
3 years ago
Read 2 more answers
Problem
Naddik [55]

Answer:

Mass of products equals mass of the reactants. If you build a campfire like this one, you start with a big pile of logs. As the fire burns, the pile of logs slowly shrinks. By the end of the evening, all that's left is a small pile of ashes.

Explanation:

i learned this in science before

5 0
3 years ago
When people smoke, carbon monoxide is released into the air. In a room of volume 70 m3, air containing 5% carbon monoxide is int
KonstantinChe [14]

Answer:

\frac {dQ}{dt}=0.0001-\frac {Q}{70}\times 0.002

Explanation:

Given that the volume is 70\ m^3

The air flows (F) at the rate of 0.002\ m^3/min. The concentration of carbon monoxide in the air is 5%.

Let Q be the amount of the carbon monoxide present in the room at time t. then,

\frac {dQ}{dt}=Rate\ of\ Q_{in}-Rate\ of\ Q_{out}

Also,

Rate\ of\ Q_{in}=5\%\ of\ 0.002=0.0001

Rate\ of\ Q_{out}=\frac {Q}{70}\times 0.002

Thus,

\frac {dQ}{dt}=0.0001-\frac {Q}{70}\times 0.002

6 0
4 years ago
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