64.92 grams of Hg(NO3)2 to make 5.00 liters of solution.

Density = mass / volume

Thus, Mass of the solution can be expressed as:

Mass of solution = Density of solution × volume of solution

Given- Density of solution = 1.126 g/ml

volume of solution = 20.0 ml

∴ Mass of 14 wt% solution of CaCl2 = (1.126 g/ml) × (20 ml)

                                                           = 22.52 g

                                                           = 22520 mg

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

. A solution that is 0.10 M NH3 and 0.10 M NH4Cl

Explanation:

A buffer consists of a weak acid and its conjugate base counterpart. HCN is a weak acid and the salt LiCN contains its counterpart conjugate base which is the cyanide ion. A buffer maintains the pH by guarding against changes in acidity or alkalinity of the solution.

A solution of ammonium chloride and ammonia will also act as a basic buffer. A buffer may also contain a weak base and its conjugate acid.

Answer:

Good buffer systems are:  

A) NH3 + NH4Cl

C) HCN + LiCN

D) HF + NaF

Explanation:

Buffers consist in a mixture of a weak acid with its salt or a weak alkaly with its salt. All buffer systems are conformed by:

1) Weak acid + salt

or

2) Weak alkaly + salt

It is very important these salts come from the weak acid or weak alkaly. It means, the anion of the acid must be the anion in the salt which is going to be part of the buffer system. On the other hand, the cation of the weak alkaly must be the cation of the salt which is going to form the salt in the buffer system.

Then, when we evaluate all options in this exercise, answers are the following:

A) 0.10 M NH3 and 0.10 M NH4Cl. It is a buffer because NH3 (ammonia) is a weak alkaly and NH4Cl is a salt coming from NH3.

Buffer component reactions:

Reaction weak alkaly:   NH3 + H2O <-----> NH4+ + OH-

Reaction salt in water:  NH4Cl ---> NH4+ + Cl-

NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Cl is a salt from NH3.

C) 0.10 M HCN and 0.10 M LiCN. It is a buffer because HCN is a weak acid and LiCN is a salt which is coming from HCN.

Buffer component reactions:

Reaction weak acid:     HCN + H2O <-----> H3O+ + CN-  

Reaction salt in water:  LiCN --> Li+ + CN-

CN- is the anion of the acid, so it must be part of the salt in the buffer system. Then LiCN is a salt from HCN.

D) 0.10 M HF and 0.10 M NaF. It is a buffer because HF is a weak acid and NaF is a salt which is coming from HF.

Buffer component reactions:

Reaction weak acid:      HF + H2O <------> H3O+ + F-

Reaction salt in water:   NaF ---> Na+ + F-

F- is the anion of the weak acid (HF), so it must be part of the salt in th buffer systema. Then NaF is a salt coming from HF.

However option B, it is not a buffer, because it is a mixture of 0.10 M HCN and 0.10 M NaF.    Salt is not coming from the weak acid.

Reaction weak acid:    HCN + H2O <-----> H3O+ + CN-  (anion of the acid is CN-)

Rection salt in water:   NaF --> Na+ + F-  (anion in the salt is F-, not CN-)

Anion of the acid is CN- and the anion in the salt is F- so it is not a salt coming from the weak acid. Then option B it is not a buffer system.

Answer:

1 litre of 1.0 M NaCl

Explanation:

When an ionic compound dissolves in water, it dissociates into ions. Consider the dissolution of sodium chloride in water;

NaCl(s) ------> Na^+(aq) + Cl^-(aq)

Hence, two solute particles are obtained from each formula unit of NaCl, a greater concentration of NaCl will contain a greater number of sodium an chloride ion particles.

Glucose is a molecular substance and does not dissociate in solution hence it yields a lesser number of particles in solution even at the same concentration as NaCl

Final answer:

The solution with the greatest number of solute particles is 1 litre of 1.0 M NaCl, as ionic compounds dissociate into individual ions, thus providing more particles per litre.

Explanation:

Given the details of the question, the solution that would be expected to contain the greatest number of solute particles would be 1 litre of 1.0 M NaCl. This is because when ionic compounds like sodium chloride are placed in water, they dissociate into individual ions. In the case of NaCl, it splits into two ions, sodium (Na+) and chloride (Cl-). Thus, a 1.0 M solution of NaCl would actually contain 2.0 moles of particles per litre because each formula unit of NaCl gives two particles. Covalently bonded molecules like glucose do not dissociate in solution, therefore, a 1.0 M glucose solution would have 1.0 mole of particles per litre.

Learn more about Solute Particles in Solutions here:

brainly.com/question/24503166

#SPJ3

Answer:

electron capture detectors (ECD)

Explanation:

Electron capture detectors (ECD) would be most sensitive for GC analysis of PCBs.

Answer: The detectors that would be most sensitive for GC analysis of PCBs are

- Electron Capture Detectors (ECD)

Explanation:

Polychlorinated biphenyls(PCBs) are oily chemicals that are manufactured(man-made) and used as lubricants in electrical equipments such as transformers. They are grouped as major environmental pollutants because they are characterised as being very stable, as they are resistant to extreme temperature and pressure. Therefore with time the build up in the environment and can cause harmful health effects.

In the environment, it can be found distributed virtually everywhere incl, air water or soil. The PCBs can be analysed in the environment through the use of gas chromatography( GC) with electron capture detector (ECD) as the most sensitive detector. The method was found to be more sensitive as the number of chlorine atoms attached to the biphenyl increases.

Here's the answer, I remember doing this problem last year.

23.5 degrees north, 77 degrees west

Answer:

0,12 μmol/L of MgF₂

Explanation:

Preparation of solutions is a common work in chemist's life.

In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in  μmol/L

You have 0,00598 μmol but not Liters.

To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:

50,0 mL (1L/1000mL) = 0,05 L of water.

Thus, concentration in  μmol/L is:

0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-

I hope it helps!