Answer:
All are having different valent cation and anion like mono,di and trivalent polyatomic ions .
A. RbNO3
B. K2S
C. NaHS
D. Mg3(PO4)2 formed by divalent Mg+2 and trivalent PO43-
E. CaHPO4
F. PbCO3 , lead is in Pb+2 form
G. SnF2
H. (NH4)2SO4
I. AgClO4
J. BCl3
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
Answer:
C. because the 4s orbital is at a lower energy level
Explanation:
The rest of the energy is passed on as food to the next level of the food chain. The figure at the left shows energy flow in a simple food chain. Notice that at each level of the food chain, about 90% of the energy is lost in the form of heat.
Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole. - R= 0.082
- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082× 353 K
Solving:
V = (4.745 moles× 0.082× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
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