Answer: Lithium and Ethanol
Explanation: As lithium and ethanol both have a density of under
1.000g/ml, they can be supported by the water. Blood and Magnesium are denser than the water and will therefore sink as the water can not support it them.
Answer:
Explanation:
The colligative properties of a given solution can be defined as the properties of that solution that are dependent on the concentration of the molecules or ions of the solute in the solution, and not on the type or identity of that solute. Examples include:
1. vapor pressure lowering
2. boiling point elevation
3. freezing point depression
4. Osmotic pressure
In this case, vapor pressure would be lowered because with an electrolyte introduced into a solution, the number of solute particles would be larger because the solute particles dissociate into ions, thereby competing with the solvent molecules at the surface of the solution, which in turn reduces the rate at which the solvent evaporates and condenses. Vapor pressure is lower compared to a solution with the same number of moles of nonelectrolyte solute.
The higher the number of ions in the solution, the greater the colligative properties of the solution will be impacted.
The presence of a strong electrolyte in solution affects the colligative properties differently than a non-electrolyte solute. Strong electrolytes dissociate into ions, increasing the number of particles in solution. This affects colligative properties such as vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.
The presence of a strong electrolyte in solution affects the colligative properties differently than the same number of moles of a non-electrolyte solute. This is because strong electrolytes dissociate into ions when dissolved in solution, while non-electrolytes do not. The dissociation of strong electrolytes increases the total number of particles in solution, which affects colligative properties such as vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.
For example, let's compare a solution of 1 mole of sodium chloride (NaCl) to a solution of 1 mole of sucrose (C12H22O11). The sodium chloride will dissociate into Na+ and Cl- ions, which means there are now 2 particles in solution (1 Na+ and 1 Cl-) instead of just 1 molecule of sucrose. This higher particle concentration will result in a greater depression of the freezing point and elevation of the boiling point compared to the sucrose solution.
In summary, the presence of a strong electrolyte increases the number of particles in solution, leading to greater deviations in colligative properties compared to the same number of moles of a non-electrolyte solute.
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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.
Substance A: 8.2 g/cm3
Substance B: 3.5 cm and 30.0g
Substance C: 10.0g and 40mL
Substance D: 0.5 g/cm3
Substance E: 2.0cm by 3.0cm by 1.0cm and 4.0g
Most Dense_ _ _ _ _
Least Dense
The order of density of substances ranging from most dense to least dense is :substance B>substance A>substance E>substance C>substance D.
It is a ratio of substance's mass per unit of volume.Symbol most commonly used for density is р.The SI unit of densityis kilogram per cubic meter .It explains how tightly a material is packed together.
There are2 types of density :1)absolute density 2) relativedensity.Absolute density is the massof any substance per unit volume and relative density is the ratio of density of a substance to the density of a given reference material.
Reference material used forrelative density is water.The instrument used for measuring density or relative density of liquids is hydrometer. Densityis measured at constant temperature and pressure.
To learn more about density and it's types click here:
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Answer:
So 1st it is B then D then E then a then C
(b) Analysis of the nitrate content of soil near a local water source. soil nitrate
(c) Measurement of the citric acid found in a lime.
Identify the following as either sample or analyte.
(1) lead
(2) paint chips
(3) soil
(4) nitrate
(5) lime wedge
(6) citric acid
Answer:
a) Analyte: lead. Sample: paint.
b) Analyte: nitrate. Sample: soil.
c) Analyte: citric acid. Sample: Lime
1) Lead: Analyte.
2) Paint chips: Sample.
3) Soil: Sample.
4) Nitrate: Analyte.
5) Lime wedge: Sample.
6) Citric acid: Analyte.
Explanation:
A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.
Thus:
a) Analyte is lead while you must take a sample of paint to analyze this lead.
b) Analyte is the nitrate while sample must be soil.
c) Analyte is citric acid and lime is the sample
1) Lead: Analyte.
2) Paint chips: Sample.
3) Soil: Sample.
4) Nitrate: Analyte.
5) Lime wedge: Sample.
6) Citric acid: Analyte.
Answer:
The specific heat of the copper is 0.771 cal/ grams °C
Explanation:
Step 1: Data given
Mass of the piece of copper = 15.0 grams
The temperature of the wire changes from 12.0 °C to 79.0 °C
The amount of heat absorbed is 775 cal
Step 2: Calculate the specific heat of copper
Q = m*c*ΔT
⇒with Q = the heat absorbed = 775 cal
⇒with m = the mass of the copper = 15.0 grams
⇒with c = the specific heat of copper = TO BE DETERMINED
⇒with ΔT = The change in temperature = 79.0 °C - 12.0 °C = 67.0 °C
775 cal = 15.0 grams * c * 67.0 °C
c = 0.771 cal/gm °C
The specific heat of the copper is 0.771 cal/ grams °C