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rusak2 [61]
3 years ago
11

Suppose a large atom bonds with a small atom.  Will the properties of the new molecule be the same as the large atom, the small

atom, or different from both?
Chemistry
1 answer:
Stels [109]3 years ago
6 0
Everything will be the same except that the atoms will be together nothing changes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
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If you have a polyatomic anion of Ammonium (NH41+), how many valence electrons must your Lewis Structure have?
tigry1 [53]

Answer:

One can draw the 3-dimensional structure of an atom once they have the Lewis Structure of an atom. The 3-dimensional geometrical structure of ammonium, NH4+ is referred to as Tetrahedral. ... But the + sign decrees that NH4+ has 8 valence shell electrons, due to the positive ion.

Explanation:

6 0
3 years ago
In the reaction CuO(s) + CO2(g) → CuCO3(s), a. CO2 is the Lewis acid and CuCO3 is its conjugate base. b. O2– acts as a Lewis bas
adoni [48]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

According to Lewis acid-base concept:

The substance which is donating electron pair is considered as Lewis base and the substance which is accepting electron pair is considered as Lewis acid.

For the given chemical reaction:

CuO(s)+CO_2(g)\rightarrow CuCO_3(s)

CO_2 is accepting electron pair and is getting converted to CO_3^{2-}. Thus, it is considered as Lewis acid.

O^{2-} present in CuO is a Lewis base because it is donating electron pair.

Thus, the correct answer is Option d.

4 0
3 years ago
For the reaction 3h2(g) + n2(g) 2nh3(g), kc = 9.0 at 350°c. calculate g° at 350°c.
miss Akunina [59]
When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° =  - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc 

and Kc = 9 

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

           = - 4536 
4 0
3 years ago
The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to
adelina 88 [10]

Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

      C₃H₈ + 5O₂  = 4 H₂O + 3CO₂      The molecular weights (g/mol) are:

MW  44    5x32      4x18    3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.

6 0
3 years ago
At room temperature, which of these are oxide minerals? Select all that apply.
DerKrebs [107]
C. and d. are oxyminerals.
7 0
3 years ago
Read 2 more answers
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