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yawa3891 [41]
3 years ago
8

What group 1 element is a NON metal?

Chemistry
1 answer:
Alecsey [184]3 years ago
7 0
Hydrogen??????????????????????
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1. Complete the balanced dissociation equation for the compound below. If the compound does not dissociate, write NR after the r
puteri [66]

Answer:

1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Explanation:

The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:

(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).

The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.

S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)

Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.

2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)

The net ionic equation is given below:

Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

7 0
2 years ago
Which of the following are effective treatments for some musculoskeletal conditions? Select all that apply. PLEASE ANSWER
Dimas [21]

Answer:

1,3, 4

Explanation:

6 0
3 years ago
Which of the following describes a compound?
Gemiola [76]

A compound is two or more chemically combined elements. Fore example NaCl, CO2, H2O

8 0
3 years ago
Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of
Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

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3 years ago
Benefits of using n2o as a refrigerant.
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3 years ago
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