If I added 27 grams of salt to water and separated the solution
with a hot plate, what amount of salt would be left behind?​

Answers

Answer 1

Answer:

The correct answer is - 27 grams.

Explanation:

salts contain atoms (ions) like Na+ and Cl, these atoms are electrically charged that leads to a big crystal as they can stick together. Those are too heavy to evaporate.

When you heat the hot plate water evaporates but the salt does not leave vessel so, the concentration of salt in the water increases. In the end, the concentration of salt gets so high and the salt will begin to recrystallize into a solid. When all of the water is gone, you will have salt left behind in the same concentration.

Thus, the correct answer is - 27 grams.


Related Questions

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 525 mL of a solution that has a concentration of Na ions of 1.10 M

Answers

Answer:

31.652g of Na3PO4

Explanation:

We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

Na3PO4 will dessicate in solution as follow:

Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e

xM Na3PO4 = (1.10 x 1)/3

xM Na3PO4 = 0.367M

Therefore, the molarity of Na3PO4 is 0.367M.

Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

Molarity of Na3PO4 = 0.367M

Volume = 525mL = 525/1000 = 0.525L

Mole of Na3PO4 =..?

Molarity = mole /Volume

0.367 = mole /0.525

Cross multiply

Mole of Na3PO4 = 0.367 x 0.525

Mole of Na3PO4 = 0.193 mole.

Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

Mole of Na3PO4 = 0.193 mole

Mass of Na3PO4 =.?

Mass = mole x molar mass

Mass of Na3PO4 = 0.193 x 164

Mass of Na3PO4 = 31.652g

Therefore, 31.652g of Na3PO4 is needed to prepare the solution.

Identify each element below, and give the symbols of the other elements in its group.

a. [Ar] 4s23d104p4
b. [Xe] 6s24f145d2
c. [Ar] 4s23d5.

Answers

Answer:

Explanation:

the electron configuration is defined as the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is

used to describe the orbitals of an atom in its ground state

The valence electrons, electrons in the outermost shell, can be used to know the chemical property

a)

Chemical Name of the Element: Selenium

Chemical Symbol: Se

Group it belong in periodic table:6A

Other Element in the same group:tellurium(Te),,sulfur(S)

atomic number = 34

Selenium is a chemical element that has symbol Se It is a nonmetal which is usually classified as metalloid with properties that are intermediate between the elements above and below in the periodic table.

b)Chemical Name of the Element:Hafnium

Chemical Symbol: Hf

Group it belong in periodic table:4B

Other Element in the same group: Titanium( Ti )Rutherfordium

atomic number: 72

Hafnium is a solid at room temperature.

c)Chemical Name of the Element: Manganese

Chemical Symbol:Mg

Group it belong in periodic table:Mn

Other Element in the same group:Bohrium(Bh) ,Technetium(Tc)

A scientist measures the standard enthalpy change for the following reaction to be -115.5 kJ: CO(g) + Cl2(g)___COCl2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of COCl2(g) is ________ kJ/mol.

Answers

Answer:

-226.0kJ = ΔH°f COCl₂(g)

Explanation:

Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.

For the reaction:

CO(g) + Cl₂(g) → COCl₂(g)

Hess's law is:

ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))

ΔH°f CO(g) is -110.5kJ/mol

ΔH°f Cl₂(g) is 0 kJ/mol

Replacing in Hess's law:

-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)

-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ

-226.0kJ = ΔH°f COCl₂(g)

How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
O A. Elements in the same group have the same number of valence
electrons.
B. The number of valence electrons increases as the atomic number
increases
C. The number of valence electrons is the same for all elements on
the periodic table.
D. Elements in the same period have the same number of valence
electrons.

Answers

I believe the answer is A

Answer:

A

Explanation:

calculate the energy in joules and calories required to heat 50.0g silver from 106c to 255c. ​

Answers

Answer:

Explanation:

use this fromula

q = m c ∆t

m  is mass of silver =50 g

∆t is difference in temperature= 255-106=149

C=  specific heat fo silver ( should be mentioned in your question )

g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass. Must show your work on scratch paper to receive credit.

Answers

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

Calculate the temperature and state the appropriate phase(s) (solid, liquid, vapor) for each substance: This temperature is: Fahrenheit A. SubstanceMelting Point (K)Boiling Point (K)Phase(s) Oxygen, O254.7590.19 Methane, CH493.15109.10 Water, H2O273.15373.15

Answers

Answer:

Explanation:

At 54.75K melting point, Oxygen is in gas (vapour) phase

At 373.15K boiling point, water is in liquid phase.

At 109.10K boiling point methane is in gas (vapour) phase.

What is an electrolyte solution?

Answers

Answer:

an electrolyte is a sustance that produce an electrically conducting solution when dissolved in a polar solvent , such as water

Explanation:

hope it helps

Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.

Answers

Question:

When 0.500 g of biphenyl (C₁₂H₁₀) undergoes combustion in a bomb calorimeter, the temperature rises from 26.8 °C to 29.5 °C. Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.

Answer;

-4870kJ/mol(3 significant figure)

s the following nuclear equation balanced? yes no

Answers

Answer:

Yes.

Explanation:

The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.

Therefore, the equation will be reduced to:

226 - 4 = 222

88 - 2 = 86

Hence, the equation is balanced.

In supersonic flights, molecules break apart and react chemically. which safety features protect the plane?

Answers

Answer:

Explanation:

The heat Shield are materials (usually made of  metals) protect us from heat by  absoring lots of heat and gradually releasing heat by surrounding  air cirucaltion

Answer:

Heat shield

Explanation: Most heat shields consist of one or more layers of stamped metal that are shaped into a shield that is designed to wrap around the exhaust manifold. The shield acts as a barrier and heat sink, preventing the heat from the manifold from reaching any of the components under the hood and potentially causing damage.

In a zero order reaction, it takes 342 seconds for 75% of a hypothetical reactant to decompose. Determine the half-life t_{1/2} in units of seconds. Do not enter units with your numerical answer. Numeric Answer:

Answers

Answer:

228 s

Explanation:

In a zero order reaction, the formula for the half life is given as;

t1/2 = [A]o / 2k

To obtain the rate constant k, we have to use;

[A] = [A]o - kt

kt = [A]o - [A]

From the question;

it takes 342 seconds for 75% of a hypothetical reactant to decompose.

We have;

t = 324

[A] = 25

[A]o = 100

Upon solving for k we have;

kt = [A]o - [A]

k = ( [A]o - [A] ) / t

k = (100 - 25 ) / 342

k = 75 / 342 = 0.2193

Solving for t1/2;

t1/2 = [A]o / 2k

t1/2 = 100 / 2(0.2193)

t1/2 = 100 / 0.4386 = 228 s

The carbon-carbon bond length in ethylene is ________ than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is ________ the HCH bond angle in ethane

Answers

Answer:

shorter

longer

Explanation:

The carbon-carbon bond length in ethylene is shorter than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is longer the HCH bond angle in ethane.

The objective of this question is to let us understand the concept of Bond Length and Bond angle among the unsaturated aliphatic hydrocarbons (i.e alkanes, alkenes and alkynes).

The variation in bond angles of  unsaturated aliphatic hydrocarbons  can be explained by two concepts; The valence shell electron pair repulsion (VSEPR) model and hybridization.

The VSEPR model determines the total number of electron pairs surrounding the central atom of a species. The total number of electron pairs  consist of the bond pairs and lone pairs. All the electron pairs( lie charge ) will then orient themselves in such a way to minimize the electrostatic repulsion between them.

As the number of the lone pairs increases from zero to 2 ; the bond angles diminish progressively.

However;

Hybridization is the mixing or blending of two or more pure atomic orbitals (s,p and d) to form two or more hybrid atomic orbitals that are identical in shape and energy . e.g sp, sp² , sp³  hybrid orbitals etc .

The shape of the geometry of this compound hence determines their bond  angle.

The shape of the geometry of ethane is tetrahedral which is 109.5° in bond angle while that of ethylene is trigonal planar which is 120°.

This is why the HCH bond angle in ethylene is longer  the HCH bond angle in ethane .

what volume at stp of (a)o2 (b)H2 (c)Cl2 gas will be liberated during electrolysis by the passage of 48250c of electricity,molar volume=22.4​

Answers

Bsbdjxjxnamizizs brush’s ebeidhcycyudejej ehducudidksne r

The volume of o₂  2.8 liters or 2.8 dm³.

The volume of H₂  2.8 liters or 5.6dm³.

The volume of H₂  2.8 liters or 5.6 dm³.

What s electrolysis?

Electrolysis is the process by which electric current is passed through an electrolyte to effect a chemical change.

Through this chemical change in which the substance loses gets oxidized or gains an electron reduced.

calculations:-

for a divalent molecule O2 4*96500, C = 1 mole of gas evolved.

          ∴48250c =48250/(4*96500)=0.125mole

volvme = 0.125*22.4= 2.8 litre.

for monovalent cl and H2 molecules, 2*96500c = 1 mole of gas is evolved.

      ∴48250c =48250/(2*96500)=0.25mole

Learn more about electrolysis here:-https://brainly.com/question/25712870

#SPJ2

Which of the following aqueous solutions are good buffer systems?
0.31 M ammonium bromide + 0.39 M ammonia
0.31 M nitrous acid + 0.25 M potassium nitrite
0.21 M perchloric acid + 0.21 M potassium perchlorate
0.16 M potassium cyanide + 0.21 M hydrocyanic acid
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite
0.13 M nitrous acid + 0.12 M potassium nitrite
0.15 M potassium hydroxide + 0.22 M potassium bromide
0.23 M hydrobromic acid + 0.20 M potassium bromide
0.34 M calcium iodide + 0.29 M potassium iodide
0.33 M ammonia + 0.30 M sodium hydroxide
0.20 M nitrous acid + 0.18 M potassium nitrite
0.30 M ammonia + 0.34 M ammonium bromide
0.29 M hydrobromic acid + 0.22 M sodium bromide
0.17 M calcium hydroxide + 0.28 M calcium bromide
0.34 M potassium iodide + 0.27 M potassium bromide

Answers

Answer:

Answers are in the explanation.

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:

0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.

0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.

0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.

0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.

0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.

0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.

0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.

0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.

0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.

0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.

0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.

0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.

0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.

0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.

0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.

475 grams of solid calcium oxide reacts with water vapor to form solid calcium hydroxide.Calculate the heat of reaction.
(A) - 221.1 kcal
(B) + 290.8 kcal
C) - 290.8 kcal
(D) + 324.6 kcal
(E) + 221.1 kcal​

Answers

Answer:

(A) - 221.1 kcal

Explanation:

Based in the reaction:

CaO(s) + H₂O(g) → Ca(OH)₂ ΔH = -109kJ/mol

When 1 mole of CaO reacts per mole of water vapor producing calcium hydroxide there are released -109kJ

475g of CaO (Molar mass CaO: 56.08g/mol) are:

475g CaO × (1mol / 56.08g) = 8.47 moles of CaO

As 1 mole of CaO in reaction release -109kJ, 8.47 moles release:

8.47 mol CaO × (-109 kJ / 1 mol CaO) = -923.2kJ are released

As 1 kCal = 4.184kJ:

-923.2kJ × (1kCal / 4.184kJ) =

-220.7kCal ≈ (A) - 221.1 kcal

What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?Cu + 2AgNO3→Cu(NO3)2+ 2Ag

Answers

Answer: 323.61 g of [tex]Ag[/tex] will be produced

Explanation:

The given balanced chemical reaction is :

[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]

According to stoichiometry :

2 moles of [tex]AgNO_3[/tex] require 1 mole of [tex]Cu[/tex]

Thus 3.00 moles of  [tex]AgNO_3[/tex] will require=[tex]\frac{1}{2}\times 3.00=1.50moles[/tex]  of [tex]Cu[/tex]

Thus [tex]AgNO_3[/tex] is the limiting reagent as it limits the formation of product.

As 2 moles of [tex]AgNO_3[/tex] give =  2 moles of [tex]Ag[/tex]

Thus 3.00 moles of [tex]AgNO_3[/tex] give =[tex]\frac{2}{2}\times 3.00=3.00moles[/tex]  of [tex]Ag[/tex]

Mass of [tex]Ag=moles\times {\text {Molar mass}}=3.00moles\times 107.87g/mol=323.61g[/tex]

Thus 323.61 g of [tex]Ag[/tex] will be produced from the given moles of both reactants.

Answer:

B). 2KNO3 Is your answer

Explanation:

Calculate the mass of NaCl required to prepare 360.0 mL of a 0.35 M solution.

Answers

Answer:

Explanation:

Molar mass of NaCl = 58.44

Molar mass of NaCl(58.44g)  addding to 1000ml give rise to 1M solution always, this can be writtne as below

58.44----- >1000 ml ---> 1M

X gram ----360 ml ---0.35 M

X Gram= 58.44*360*0.35/1000*1 = 7.3 gram

hope you understood the solution

Which diagram represents the bonding pattern of metals?

Answers

Answer:

there's no image can't help without it sorry

A proposed mechanism for the reaction of NO2 and CO is

Step 1: Slow, endothermic:
2 NO2 (g) → NO (g) + NO3 (g)

Step 2: Fast, exothermic:
NO3 (g) + CO (g) → NO2 (g) + CO2 (g)

Overall reaction, exothermic:

NO2 (g) + CO (g) → NO (g) + CO2 (g)

a. Identify each of the following as a reactant, product, or intermediate:
1. NO2(g)
2. CO(g)
3. NO3(g)
4. CO2(g)
5. NO(g)

b. Draw a reaction coordinate diagram for this reaction. Indicate on this drawing the activation energy for each step and the overall enthalpy change.

Answers

Answer:

a.

1. NO2(g): Reactant.

2. CO(g): Reactant.

3. NO3(g): Intermediate.

4. CO2(g): Product.

5. NO(g): Product.

b. See attached picture.

Explanation:

Hello,

a. In this case, given the reactions, we can identify each species as:

1. NO2(g): Reactant because it remains at the left side in the overall reaction.

2. CO(g): Reactant because it remains at the left side in the overall reaction.

3. NO3(g): Intermediate because it is a product in the step 1 and a reactant in step 2, for that reason it is not present in the overall reaction.

4. CO2(g): Product because it remains at the right side in the overall reaction.

5. NO(g): Product because it remains at the right side in the overall reaction.

b. In this case, given that the first step is slow and endothermic, it has a high activation energy and the products will have more energy than the reactants, for that reason the final energy is above the initial point. Moreover, since the second step is fast and exothermic, it has a low activation energy and the products will have less energy than the reactants, for that reason, the reaction coordinate diagram is shown on the attached file.

Regards.

Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set-up

Answers

The question is incomplete; the complete question is;

Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set- up? Select one

a) There was an air bubble in the burette tip.

b) There is not enough indicator in the analyte.

c) The burette tip is leaking titrant into the analyte.

d) The analyte solution is being stirred too quickly

Answer:

a) There was an air bubble in the burette tip.

Explanation:

Titration involves the determination of the concentration of a solution by measuring the volumes of reactants used in the reaction. The concentration of one of the species must be known while the concentration of the other specie is to be determined by the volumetric analysis.

However, if there are air bubbles at the tip of the burette, this will cause less volume of titrant to be delivered from the burette than expected. Hence, the analyst may think that a certain volume of titrant has been delivered while in reality, a lesser volume was actually delivered due to the air bubbles present. Hence, the analyst may pass the expected endpoint without any colour change because of this problem.

From the available options to the question:

a) There was an air bubble in the burette tip.

b) There is not enough indicator in the analyte.

c) The burette tip is leaking titrant into the analyte.

d) The analyte solution is being stirred too quickly

The most likely problem with the titration setup that could make one to pass the expected endpoint with no visible color change would be if there is not enough indicator in the analyte. The correct option would be B.

A suitable quantity (in drops) of the indicator should be added to the analyte in the conical flask before carrying out a titration. The color of indicators changes quickly near their pKa.

If too few drops of the indicator is used, the color change will be too faint to be obvious and the endpoint will be exceeded. If too many drops of the indicator is used, the final pH of the reaction would be affected and the titer value will be inaccurate.

In this case, the expected endpoint has been exceeded without any color change. The most likely problem would, therefore, be that there is not enough indicator in the analyte.

More on indicators can be found here: https://brainly.com/question/4050911

Mathematical relationship between the rate of a chemical reaction and the
concentration of reactants?

Specific Rate Constant

Collision Theory

Rate Law

Activation energy

Answers

Answer:

Specific Rate Constant

Explanation:

Mathematical relationship between the rate of a chemical reaction and the

concentration of reactants is Specific Rate Constant.

how to use VBT to draw the orbital overlapping diagram ​

Answers

Answer:

from the valence elecrtons configuration is the centre atom.atomised the number of elecrtons pair determine the hybridization .

Explanation:

you can read this note to know the ans

An electron in a 3s3s orbital penetrates into the region occupied by core electrons more than electrons in a 3p3p orbital. An electron in a orbital penetrates into the region occupied by core electrons more than electrons in a orbital. true false

Answers

Answer:

True

Explanation:

The penetrating ability of electrons in the orbitals is in the order s > p > d > f

An electron in a 3s orbital is closer to the nucleus than the one in a 3p orbital and as a result, there will be lesser shielding effect on it. This low shielding effect experienced by the 3s electron gives it a high penetration ability and hence will be able to easily penetrate regions occupied by core electrons. Conversely, the 3p orbital is farther away from the nucleus, electrons revolving around it are highly shielded which limits their ability to penetrate regions of core electrons.

Note that the maximum electrons that the s orbital can accommodate is 2 while p orbital can accommodate a maximum of 8.

Which describes an effect that ocean currents have on short-term climate change? Ocean currents increase the strength of prevailing winds, which can cool the air and land. Ocean currents can carry cold water, which can cool the air and land. Ocean currents increase hurricane activity, which can raise the temperature of the air and land. Ocean currents can carry warm water, which causes hurricane activity and raises the temperature of the air and land.

Answers

Answer:  B

Ocean currents can carry cold water, which can cool the air and land.

Explanation:

eet ees wat eet ees

plz mark brainliest

Answer:

B is right

Explanation:

which proess is part of the carbon cycle

Answers

Answer:

The key processes in the carbon cycle are: carbon dioxide from the atmosphere is converted into plant material in the biosphere by photosynthesis.

Explanation:

organisms in the biosphere obtain energy by respiration and so release carbon dioxide that was originally trapped by photosynthesis. ... The carbon becomes part of the .

In a chemical equation, which symbol should be used to indicate that a substance is in solution? (s)

Answers

Answer:

(aq) meaning aqueous solution

Explanation:

hope it helps .

Which is regarding enzyme inhbition

Answers

Explanation:

An enzyme inhibitor is a molecule that binds to an enzyme and decreases its activity. ... Since blocking an enzyme's activity can kill a pathogen or correct a metabolic imbalance, many drugs are enzyme inhibitors. They are also used in pesticides.

a) What is the common-ion effect?b) Given an example of a salt that can decrease the ionization of HNO2 in solution. List all of the substances (species) that would be present in the solution. (Consider the HNO2, the salt, and the water.)

Answers

Answer:

Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.

NaNO2

Explanation:

In order to understand exactly what common ion effect is, let us consider a simple unambiguous example. Assuming I have a solution of an ionic substance that contains a cation A and an anion B, this ionic substance has chemical formula AB. Secondly, I have another ionic distance with cation C and anion B, its chemical formula is CB. Both CB and AB are soluble in water to a certain degree as shown by their respective KSp.

If I dissolve AB in water and form a solution, subsequently, I add solid CB to this solution, the solubility of CB in this solution is found to be lees than the solubility of CB in pure water because of the ion B^- which is common to both substances in solution. We refer to the phenomenon described above as common ion effect.

Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.

If I try to dissolve NaNO2 in a solution of HNO2, the solubility of NaNO2 in the HNO2 solution will be less than its solubility in pure water due to common ion effect. Also, the extent of ionization of HNO2 in a system that already contains NaNO2 will be decreased compared to its extent ionization in pure water. This system described here will contain HNO2, water and NaNO2

Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass of 10.88 g is added and the volume of the water now reads 49.4 mL. What is the density of the metal in g/mL? Multiple Choice 10.9 1.8 0.250 1.6 0.541

Answers

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

Final volume = 49.4ml

Mass = 10.88g

Density = ?

Volume = Final volume - initial volume

= 49.4 - 43.5

= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

= 1.8g/ml

Other Questions
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