Answer:
1)Carbonated water is saturated with carbon, hence it gives off carbon through bubbles.
2)Adding sugar to water until it no longer dissolves creates a saturated solution.
3)Continuing to dissolve salt in water until it will no longer dissolve creates a saturated solution.
An unsaturated tea and sugar solution would be one into which you could add more sugar and have the sugar still dissolve
First, we write the reaction equation:
3Pb(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃(PO₄)₂
Moles of Pb ions present:
moles = concentration x volume
= 0.15 x 0.25
= 0.0375
From the equation,
moles Pb : moles Na₃PO₄
= 3 : 2
Moles of Na₃PO₄:
2/3 x 0.0375
= 0.025
volume = moles / concentration
= 0.025 / 0.1
= 0.25 L
= 250 ml
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
250 kJ/87.9 KJ per mole Cl2 * 71g/mole= 202 g It is D for plato users
Answer:
Following are the answer to this question:
Explanation:
In the given question information is missing, that is equation which can be defined as follows:

- Growing temperatures may change its connection to just the way which consumes thermal energy in accordance with Le chatelier concepts Potential connection is endothermic. Answer: shifts to the right
-
Kc are described as a related to the concentration by the intensity of both the reaction for each phrase which reaches a power equal towards its stoichiometric equation coefficient Kc = \frac{product}{reactant}
It increases [product] but reduces [reactant] Therefore, Kc increases