If you’re asking to balance the equation then:
Pb(NO3)2(aq) + 2KCl(aq) -> 2KNO3(aq) + PbCl2(s)
Just remember: the equations at the end is Cl not C12
Note: the small number on the bottom (subscripts) apply to the one element if it’s inside the bracket and if the small number is on the outside of the bracket it applies to all the elements. For example the 3 in (NO3)2 applied only to the O (oxygen) and the 2 applies to both N and O but don’t forget it’s multiplied. So it would be 2 N’s and 6 O’s bc the 3 multiplies with the 2 only for the O.
Answer:
physical
Explanation:
no chemical reaction is happening
If you don't practice enough it's obviously going to be hard but if you practice enough it's going to be a piece of cake so don't think if it's going to be hard or not just think it's going to be worth the try at the very end
Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
Answer:
A collapse of the population is rotting, food is not enough and livelihoods have become unfeasible to decrease the number of individuals again.
Another way is to generate mutations to generate a species more vulnerable to decreasing numbers.
In this way the overpopulation is controlled.
Explanation:
In ecosystems, if an increased population breaks the balance of this and begins a new constant adaptation of the extinction of some and overpopulation of others, which may be some chains break or remain unstable.
