The ions formed are NH4(+) and S(2-)
The dissolution reaction of (NH4) 2S in water is as follows:
(NH4) 2S ==> 2 NH4 (+) + S (2-).
Ammonium sulfide is the ammonium salt of hydrogen sulfide. It has the formula (NH4) 2S and belongs to the sulfide family.
It is a relatively unstable compound (crystals decomposing at -18 ° C, but exists and is more stable in aqueous solution.) With a pKa exceeding 15, the hydrosulfide ion cannot be significantly deprotonated by ammonia. Thus, such solutions consist mainly of a mixture of ammonia and hydrosulphide of ammonium, it has a smell, close to that of hydrogen sulfide, and its aqueous solutions can be precisely by emitting H2S.
<u>Answer:</u>
<u>Plasmas of great interest to scientists or manufacturers as</u>
- Plasma is electrically charged gases that contain considerable charged particles that can change the behavior of the substance.
<u>Current uses of plasmas:</u>
- First, it is used to make semiconductors for different types of electronic equipment
- Secondly, they're used in making transmitters for high-temperature films.
<u>Way scientists and engineers hope to use plasmas in the future:</u>
- The scientists are hoping to use plasma in the future to get rid of all hazardous wastes through a process called plasma gasification.
Answer:
4.) 9, 1, and 4 5.) 4, 1, and 4
Explanation:
I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:
number 4
H: 9
P: 1
O: 4
number 5:
H: 4
S: 1
O: 4
you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.
hope this helped you :)
When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
Learn more about balancing equation here:
brainly.com/question/26227625
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Delta H of solution = -Lattice Energy + Hydration
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>
<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
<span>multiply moles to the delta H of solution to cross cancel moles. .75moles x -64kJ/mol =4.7</span>
