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belka [17]
3 years ago
7

Given the chemical formula, how is this substance MOST LIKELY classified? A) atom B) compound C) both atom and compound D) neith

er atom nor compound
Chemistry
1 answer:
cluponka [151]3 years ago
6 0

Answer:

The answer s actually B

Explanation:

The answer is B because usatestprep said so.

Hope this helps

Do not trust the other answer the correct one is BBBBBBBBBBBB

B) Compound

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Because copper is a metal, it is what?
oksano4ka [1.4K]

Copper is a mineral and an element essential to our everyday lives. It is a major industrial metal because of its high ductility, malleability, thermal and electrical conductivity and resistance to corrosion. It is an essential nutrient in our daily diet. And, its antimicrobial property is becoming increasingly important to the prevention of infection. It ranks third after iron and aluminum in terms of quantities consumed in the USA.


6 0
3 years ago
True or False: When you measure something with a ruler, you should NOT estimate an extra digit. You only need to report the numb
sveticcg [70]

Answer:

False

Explanation:

It is important to give exact measurements.

5 0
3 years ago
What must occur in order for a chemical reaction to be specifically classified as endothermic instead of exothermic?
Alla [95]

Answer:  B- Chemical bonds are formed. Energy is released in the form of heat.

Explanation: I hoped that helped !

6 0
3 years ago
Read 2 more answers
How many oxygen atoms are contained in 2.74 g of Al2(SO4)3?
abruzzese [7]

Answer:

5.79 × 10^23 Oxygen atoms

Explanation:

Number of Oxygen atom in the compound = 4×3 = 12

Molar mass of Al2(SO4)3 = 342 g/mol.

No of mole = mass/molar mass = 2.74/342 = 8.01×10^-03 mole

2.74g of Al2(SO4)3 × 1 mole of Al2 (SO4)3 / 342g of Al2 (SO4)3 * 12 mole of Oxygen/ 1mole of Al(SO4)3 * 6.02×10^23 Oxygen atom/ 1 mole of Oxygen

= 5.79×10^23 Oxygen atoms

3 0
3 years ago
Read 2 more answers
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
sveta [45]

Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

4 0
3 years ago
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