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katovenus [111]
3 years ago
8

Write the balcanced molecular and net ionic equation for the acid-base reaction:

Chemistry
1 answer:
poizon [28]3 years ago
7 0

Explanation: For the given acid-base reaction, The molecular equation is:

Molecular equation: HNO_3(aq.)+CH_3NH_2(aq.)\rightarrow CH_3NH_3^+(aq.)+NO_3^-(aq.)

As we know, HNO_3 is a strong acid, so in the ionic equation, it will completely dissociate into its respective ions.

The ionic equation for the given acid-base reaction is given by:

H^+(aq.)+NO_3^-(aq.)+CH_3NH_2(aq.)\rightarrow CH_3NH_3^+(aq.)+NO_3^-(aq.)

As nitrate ions are present on both the sides, so in the net ionic equation, it will get cancelled out, then the net ionic equation becomes:

Net ionic equation: H^+(aq.)+CH_3NH_2(aq.)\rightarrow CH_3NH_3^+(aq.)

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The oxidation of at least two atoms should change
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How many electrons are there in an atom of nickel?.
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Nickel has 28 electrons
5 0
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what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?
MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now  F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻  +  H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF,  the [ OH ⁻] from the given pH, and finally the mass needed to produce that  OH⁻ concentration.  

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is  6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14  ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

                                   F⁻                    HF                        OH⁻

Equilibrium                 X                  2.51 x 10⁻⁶            2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X  =  1.5 x 10⁻¹¹     ⇒  X =  (2.51 x 10⁻⁶)²  / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L  *  0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF *  42.0 g/mol NaF = 6.03 g NaF

7 0
3 years ago
“True or False”
Sliva [168]
True, I’m sorry if I’m wrong
3 0
3 years ago
The following substances dissolve when added to water. Classify the substances according to the strongest solute-solvent interac
Norma-Jean [14]

Answer:

CuCl2-Ion-dipole forces

CuSO4-Ion-dipole forces

NH3-Dipole-dipole forces

CH3OH-Dipole-dipole forces

Explanation:

Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.

Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.

Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles

3 0
3 years ago
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