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Ilia_Sergeevich [38]
3 years ago
8

Carbon containing compounds are referred to as:

Chemistry
1 answer:
Flura [38]3 years ago
3 0
Organic compounds
If you need an explanation I can give one!
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PLEASE HELP ASAP
sergij07 [2.7K]
Both liquids and gasses are fluids is the answer
7 0
3 years ago
29. The chemical equation H2 + O2 + H2O represents what type of equation?
slavikrds [6]

Answer:

The chemical equation H2 + O2 + H2O represents what type of equation?

A. A balanced equation

xXxAnimexXX

4 0
2 years ago
Which of the following is an example of ionic compound?
spayn [35]

What you are looking for is something from the left side of the periodic table (the metals), combined with something from the right side of the periodic table.

SiCl4 is something from the middle with something on the right. Not the answer.

HCl is a possible answer, but it is not the best one, because Hydrogen can be on both sides. It is not quite as willing to give up its electrons as the answer.

CCl4 has the same problem as A.

The answer is C

Ca is in column 2 just about as far left as you can get.

Cl is in column 17 which is just about as far right as you can get.

5 0
3 years ago
Where is the color rock located?
ludmilkaskok [199]
Ans: The fluorescent-colored rock stacks rise from the desert floor more than 20 miles south of the Las Vegas Strip. Crafted from local limestone, the boulders were painted in fluorescent hues before being stacked. The towers are between 25 and 35 feet tall.
3 0
3 years ago
If the rate of disappearance of N2O5 is equal to 1.40 mol/min at a particular moment, what is the rate of appearance of NO2 at t
kramer

Answer:

2.8 mol/min  is the rate of appearance of NO_2 at that moment.

Explanation:

2N_2O_5\rightarrow 4NO_2+O_2

Rate of the reaction = R

R=\frac{-1}{2}\frac{d[N_2O_5]}{dt}=\frac{1}{4}\frac{d[NO_2]}{dt}=\frac{1}{1}\frac{d[O_2]}{dt}

Rate of disappearance of N_2O_5 is =-\frac{d[N_2O_5]}{dt}= 1.40 mol/min

R=\frac{-1}{2}\frac{d[NO_2]}{dt}

R=\frac{1}{2}\times 1.40 mol/min=0.70 mol/min

Rate of appearance of NO_2 is =\frac{1}{4}\frac{d[NO_2]}{dt}:

R=\frac{1}{4}\frac{d[NO_2]}{dt}

\frac{d[NO_2]}{dt}=4\times 0.70 mol/min=2.8 mol/min

2.8 mol/min  is the rate of appearance of NO_2 at that moment.

7 0
3 years ago
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