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lutik1710 [3]
3 years ago
6

8. Which chemical reaction involves the fewest Hydrogen,

Chemistry
1 answer:
pochemuha3 years ago
8 0

Answer:

c

Explanation:

if ya look at it it seems to have the least amount of H's

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Write the cations and anions present in NaCl​
pochemuha

In table salt, sodium chloride, sodium is the cation (Na+) and chloride is the anion (Cl-).
6 0
2 years ago
The repeating subunits that are responsible for the chance of a crystal are know as
Naddik [55]
Unit cells.
They are the smallest group of atoms that take form of a crystal and can repeat in 3 dimensions.
7 0
3 years ago
How many types of hybridization does carbon undergo?​
Novosadov [1.4K]

Answer:

For carbon the most important forms of hybridization are the sp2- and sp3- hybridization. Besides these structures there are more possiblities to mix dif- ferent molecular orbitals to a hybrid orbital. An important one is the sp- hybridization, where one s- and one p-orbital are mixed together.

3 0
2 years ago
The density of aluminum is 2.70 g/cm3. What is the mass of a solid piece of aluminum with a volume of 2.50 cm3?
blagie [28]

Answer:

<h2>6.75 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 2.7 × 2.5

We have the final answer as

<h3>6.75 g</h3>

Hope this helps you

4 0
2 years ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-cell reactions are:

Oxidation half reaction (anode):  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction (cathode):  Pb^{2+}+2e^-\rightarrow Pb

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 25^oC=273+25=298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +0.63 V

E_{cell} = cell potential for the reaction = ?

[Zn^{2+}] = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now put all the given values in the above equation, we get:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

Therefore, the cell potential for this reaction is 0.50 V

5 0
3 years ago
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