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Molodets [167]
3 years ago
7

Which is an example of covalently bonded atoms?

Chemistry
2 answers:
Marrrta [24]3 years ago
6 0
C) Cl2 there hope this helps people
pychu [463]3 years ago
4 0

Answer:

C) Cl2

Explanation:

Covalent bonds are when atoms are non-metal. No metals are involved.

By referencing the periodic table, we can determine their columns or groups, which shows their families and whether each of these elements are metal or non-metal.

A) Be is berylium, in the second column (group 2A), which are Alkaline Earth Metals. Since Be is a metal, BeCl2 cannot be covalent.

B) Ca is Calcium, in the second column (group 2A), which are Alkaline Earth Metals Since Ca is a metal, CaCl2 cannot be covalent.

C) The only atom is Cl, which is chlorine. It is in the 17th group or group 7A. These are halogens, a type of non-metal. This can be covalent because there are only non-metal.

D) Ti is titanium. It's in the middle section, the transition metals. Because its a metal, TiN cannot be covalent.

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How many total atoms are in 0.380 g of P2O5
pogonyaev
The gram formula mass of P2O5 is 31*2 + 16*5 =142. The mole number of P2O5 is 0.380/142=0.00268 mol. There is 7 atom in one P2O5 molecule. And there is 6.02 * 10^23 molecule per mol. So the answer is 1.13*10^22 atoms.
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Help please!!!!!!!!!
Hatshy [7]

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D. Element

Explanation:

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3 0
3 years ago
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:2NH3(g) + 3I2(g) ⇌ N2(g) + 6
miss Akunina [59]

Answer: The value of K_{c} for this reaction is 250000.

Explanation:

The given equation is as follows.

2NH_{3}(g) + 3I_{2}(g) \rightleftharpoons N_{2}(g) + 6HI(g)

N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); K_{c_{1}} = 0.50   ... (1)

H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g); K_{c_{2}} = 50  ... (2)

To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.

3H_{2}(g) + 3I_{2}(g) \rightleftharpoons 6HI(g); K_{c_{2}} = (50)^{3}  ... (3)

Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.

3I_{2} - N_{2} \rightleftharpoons 6HI - 2NH_{3}

This equation can also be re-written as follows.

3I_{2} + 2NH_{3} \rightleftharpoons N_{2} + 6HI

This equation is similar to the equilibrium equation given to us.

Therefore, during this subtraction the equation constants get divided as follows.

K^{'}_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\\= \frac{(50)^{3}}{0.50}\\= 250000

Thus, we can conclude that the value of K_{c} for this reaction is 250000.

6 0
3 years ago
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