The concentration of a solution of sodium hydroxide is 0.10 mol/dm3. What is the concentration of the solution in g/dm3?​

Answer:

All forms of hydroxides

Explanation:

The metals in Group IA (Li, Na, K, Rb, Cs, and Fr) are called alkali metals because they all form hydroxides (such as NaOH) that were once known as alkalies. The electron configurations of the alkali metals are characterized by a single valence electron.

Sleet ,freezing of raindrops, snowflakes Precipitation

are 4 non examples of condensation .

What is condensation ?

Condensation, the separation of a liquid or solid from a vapor, generally occurs on surfaces that are cooler than the adjacent gas. A substance condenses when the pressure exerted by that vapor exceeds the vapor pressure of the substance's liquid or solid phase at the temperature of the surface on which condensation is occurring. As the steam condenses, heat is released. If this heat is not removed, the surface temperature will rise until it matches the surrounding steam.

If the air does not contain small particles, so-called aerosols, condensation will only occur if the air is extremely supersaturated with water vapor.

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Carbonated water is used makes sprite a substitute for egg in baking

Carbonated water can add moisture to recipes, but it also works as a great leavening agent

What is carbonated water ?

Carbonated water (soda water, bottled water, sparkling water, club soda, gassed water, also known as bottled water in many places, especially in the United States as seltzer or seltzer water) contains dissolved carbon dioxide. It is also water containing gas. It is artificially injected under pressure or occurs by natural geological processes. Carbonation creates tiny bubbles, giving the water a carbonation quality. Common forms include sparkling natural mineral water, club soda, and commercial sparkling water.

Carbonated water can add moisture to recipes, but it also works as a great leavening agent. Carbonation traps air bubbles and makes the finished product light and fluffy. You can substitute 60 grams of sparkling water per egg. This alternative is great for cakes, cupcakes and quick breads

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0.021 moles of KCN are required, to prepare a buffer with a pH of 9.230

A buffer, an aqueous solution consists of a mixture of a weak acid and its conjugate base or vice versa. Addition of a small amount of strong acid or strong base hardly changes the pH.

Given,

Volume = 125 mL

Concentration of HCN = 0.420 M

pH = 9.230

Moles of HCN = Volume × Concentration

= 0.125 × 0.420

= 0.0525 moles

HCN + KOH ⇄ KCN + H₂O

0.0525 + X   ⇄    0                     [initially]

(0.0525 - X) + 0  ⇄ X                  [At equilibrium]

pH = pKₐ + log [salt]/[acid]

9.230 = 9.31 + log [salt]/[acid]

-0.08 = log [salt]/[acid]

0.83 = [salt]/[acid]

X / (0.0525 - X) = 0.83

X = 0.021 moles of KCN

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Alkali carbonates are soluble in water so (NH₄)2CO₃ sodium carbonate and potassium carbonates are soluble in water.

A composite is a material comprised of two or more constituent parts. Examples of molecules are carbon dioxide, freshwater, and table salt. a substance produced by the chemical joining of two or three dissimilar components. Examples of compounds are salt (NaCl), which is made up of chloride and sodium atoms, and water (H2O), which would be made up of hydrogen and oxygen atoms.

When a group of atoms unite to form a pure material, a compound is created. These can be reduced to their individual atoms. Atoms from the reactants interact during chemical reactions to create new compounds. Reactants are the species that are reacting, while products are the results of the reaction.

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Iron bromide will form principate during the reaction.

When cations as well as anions in an aqueous solution combine to generate a precipitate, an insoluble solid, precipitation processes take place.

When target ion concentrations were identical, the ion that forms the poorest soluble compound would precipitate initially (at the lowest counter ion concentration), followed by the other ions as when the solubilities of respective compounds are attained.

When fe2(co3)3 reacts with mgbr2 then it will form iron bromide and magnesium carbonate.

Its chemical reaction can be written as:

[tex]Fe_{2}CO_{3} +MgBr_{2}[/tex] →[tex]FeBr_{3} +MgCO_{3}[/tex]

Here, magnesium carbonate will form ppt because it it will be a insoluble.

That's why it can be said that it will form ppt during the reaction.

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The diode is forward-biased. The acceptor concentration on the p-side is [tex]N_{A}[/tex] = 10¹⁶ cm⁻³. The donor concentration on the n-side is [tex]N_{D}[/tex] = 10¹⁴ cm⁻³.  The intrinsic carrier concentration is [tex]n_{i}[/tex] = 3.16×10¹¹ cm⁻³. Yes, the low-level injection condition applies. The bias (in volts) applied to the diode is [tex]V_{A}[/tex] = 0.18V.

The diode is forward-biased because of the excess electrons on the P-side and the excess holes on the N-side. In the intrinsic carrier concentration on the P-side n₀p₀ = 10¹⁶ × 10⁷ = 10²³ = [tex]n_{i}[/tex] = √10²³ = 3.16×10¹¹ cm⁻³, while on the N-side n₀p₀ = 10¹⁴×10⁹ = [tex]n_{i}[/tex] = √10²³ = 3.16×10¹¹. The low-level injection applies on the P-side as Δn(-[tex]x_{p}[/tex]) = 10¹⁰ << p₀ and on the N-side Δp([tex]x_{n}[/tex])=10¹² << n₀ = 10¹⁴.

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Assume at the end of step b in “The Copper Cycle” experiment, we had only solid product b and water. Considering the reactions after that, the ions present in the water before the rinsing and decanting are: copper (II) ions and hydroxide ions.

The copper cycle experiment aims to perform a series of reactions that regenerate elemental copper at the end of the copper cycle. In this cycle, the reaction series includes single replacement, double replacement, synthesis, and decomposition reactions.

In step b, the ions that exist in the water are copper (II) and hydroxide ions which are stated as: [Cu(H2O)6]2+(aq) + 2OH- --> Cu(OH)2(s) + 6H2O(l).

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The percent difference between the pressure calculated using the van der waals' equation and the ideal pressure is 5.89 %.

This can be calculated using van der waals' equation, followed by calculating pressure difference by: [P(ideal)-P(Van der waal's equation)/P(ideal)]*100

P(ideal) is given: 16.45

P(van der waals'): [tex](P+n2aV2)(V−nb)=nRT[/tex]

T= 270.9K  

V=1.301 L

a for oxygen = 1.36 (L2 atm mol-2)

b for oxygen =  0.0319 (L mol-1)

n= 0.9686 mol

Using equation P(van der waals's) = 15.48 bar

and difference is 5.81%

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The homozygous dominant genotype is 0%, the heterozygous genotype is 50%, and the homozygous recessive genotype is 50%.

This problem does not display the Punnett box. If the genotype of the heterozygous parent has spots (Ff) and the other parent is homozygous without spots, the cross between the two parents is:

Parental: Ff >< ff

Gametes: F . f >< f

Filial: Ff (heterozygous having freckles) and ff (recessive homozygous over not having freckles)

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Answer: 0%, 50%, 50%

Explanation:

The impedance of an AC circuit with 10 ohm resistor, 12 ohm inductive reactance and 7 ohm capacitive reactance is 11.2 ohm.

What is impedance?

Impedance is the resistance of a circuit to AC which is caused by the combining effects of ohmic resistance and reactance. It is represented by the symbol Z and its unit is ohm. It is a measure of opposition to electric flow.

The impedance of an AC circuit can be calculated by,

Z = [(R^2) + (XL - XC)^2]^1/2

where R = resistance, XL = inductive reactance, XC = capacitive reactance

Z = [(10^2) + (12 - 7)^2]^1/2 = (100 + 25)^1/2 = 11.2 ohm.

Therefore, the impedance of an AC circuit with 10 Ohm resistor, 12 ohm inductive reactance and 7 ohm capacitive reactance is 11.2 ohm.

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Option A, In the polysaccharide cellulose, large numbers of monosaccharides are linked together through glycosidic linkages.

The most prevalent carbohydrates in nature are polysaccharides, which have several uses including energy storage and inclusion in plant cell walls. Large polymers known as polysaccharides are created by joining tens to thousands of monosaccharides together via glycosidic connections. Starch, glycogen, and cellulose are the three polysaccharides that are most prevalent. Because each of these three produces only one kind of monosaccharide (glucose) after full hydrolysis, they are collectively known as homopolymers. In addition to monosaccharides, heteropolymers may also contain sugar acids, amino sugars, or noncarbohydrate compounds. Non-reducing carbohydrates known as polysaccharides lack the sweetness and do not mutate.

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AlCl3 is the answer.

As strong acid-strong base salts, NaCl and MgCl2 are both neutral.

Since NH4Cl is a weak base-strong acid salt, it is also acidic. AlCl3 is, nevertheless, a stronger acid than NH4Cl.

Neither acids nor bases can be found in neutral substances. Accordingly, neutral substances have an equal number of these ions. An acid has hydrogen ions, a base has hydroxyl ions, etc. Water is the most typical example of a neutral substance. Both acidic and basic properties are absent from water.

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The pressure inside the vessel at temperature of 35 degree Celsius is 0.51atm.

According to the ideal gas equation,

PV = nRT

Where,

P is the pressure of the vessel,

V is the volume of the vessel which is 2.9 L in this case,

n is the number of moles of gas,

R is the gas constant,

T is the temperature of the gas.

n = 2.6/44

R = 0.082 m³-atm/kmol-K

Temperature = (35+273) K.

Temperature = 308° K

Putting the values,

P = 2.6/44 x 0.082 x 308/2.9

P = 0.51 atm.

The pressure of the vessel is 0.51 atm.

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A 40.0 ml sample of 0.25 m KOH is added to 60.0 ml of 0.15 m Ba(OH)₂.  the molar concentration of OH⁻ ion in the resulting solution is 0.28 M.

Given that :

volume of KOH = 40 mL = 0.04 L

Molarity of KOH = 0.25 M

volume of  Ba(OH)₂ = 60 mL = 0.06 L

molarity of  Ba(OH)₂ = 0.15 M

moles of OH⁻ in KOH = 0.04 × 0.25 = 0.01

moles of OH⁻ in Ba(OH)₂ = 2 (0.06 × 0.15) = 0.018

total no. of moles = 0.01 + 0.018 = 0.028 mol

total volume = 0.04 + 0.06 = 0.1 L

molarity of OH⁻ ion = moles / volume

                                 = 0.028 / 0.1

                                 = 0.28 M

Thus, A 40.0 ml sample of 0.25 m KOH is added to 60.0 ml of 0.15 m Ba(OH)₂.  the molar concentration of OH⁻ ion in the resulting solution is 0.28 M.

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melting of gold, burning of magnesium, and water evaporation.

One instance of a physical change is melting. A physical change is when a sample of matter experiences a change in some of its qualities but not in its identity. Water turns into water vapor when it is heated.

Physical changes, such as melting, evaporation, and condensation, are different from those that result in the synthesis of new materials through chemical reactions.

Magnesium reacts with water to produce hydrogen gas, which, when exposed to the intense heat and oxygen present in air, ignites explosively. When it reacts, the heat generated causes the water molecule to shrink and transform into hydrogen gas.

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As we move across period 3 the number of delocalized electrons per metal atom increases and the radius of the elements decreases. This means the melting point increases.

What is melting point and boiling point ?

The temperature at which molecules in a solid can pass one another and transform into liquids is known as the melting point. On the other hand, liquids and gases are involved in the boiling point. Some molecules at the liquid's surface are escaping as the molecules move around.

The three metals' melting and boiling points rise as a result of the strengthening of their metallic connections. Each atom can contribute an increasing quantity of electrons to the delocalized "sea of electrons." As you get from sodium to magnesium to aluminum, the atoms also get smaller and have more protons.

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The volume of a 2.0 m solution of NaOH needed to prepare 494.4 ml of a 0.15 m solution of NaOH is 37.08 ml.

Dilution is defined as the process in which the concentration of a sample is decreased by adding more solvent. The dilution formula is given below.

C₁V₁ = C₂V₂

where C₁ = initial concentration of sample = 2.0 m

V₁ = initial volume of sample

C₂ = final concentration after dilution = 0.15 m

V₂ = total final volume after dilution = 494.4 ml

Plug in the values to the formula and solve for the volume of the solution needed.

V₁ = C₂V₂/C₁

V₁ = (0.15 m)(494.4 ml)/(2.0 m)

V₁ = 37.08 ml

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Option 4. nutrients (nitrate, phosphate, silica) are non-conservative properties of a deep water mass.

The non-conservative factors like phosphorous (P), iron (Fe), and carbon (C) or dissolved inorganic carbon found in seawater are chemically or biologically reactive. Conservative elements have longer residence instances. that is the purpose these factors make up the majority of the sea's dissolved materials.

Sodium and chloride are nonconservative constituents of seawater. Non-conservative pollutants approach pollutants that are presumed to be destroyed, biodegraded, chemically converted, or volatilized within the POTW to some diploma.

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1.81 liters of carbon dioxide are produced when 5.00 grams of hydrogen bicarbonate breaks down.

Mass of hydrogen bicarbonate (H₂CO₃) = 5.00 g

Molar mass of hydrogen bicarbonate = 62 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Density of CO₂ = 1.97 g / L

Temperature = 25 0C

Pressure = 1.24 atm

Volume of CO₂ = ?

Write the balanced chemical equation

    H₂CO₃ →  CO₂ + H₂O

The molar ratio of hydrogen bicarbonate to carbon dioxide is 1 : 1

It means that when one mole of hydrogen bicarbonate decomposes it will produce one mole of carbon dioxide

Now we will find out the number of moles of hydrogen bicarbonate

    number of moles = mass / molar mass

    number of moles of H₂CO₃ = 5.00 g / 62 g/mol

    number of moles of H₂CO₃ = 0.081 mol

Number of moles of H₂CO₃ = number of moles of CO₂

Hence the number of moles of CO₂ = 0.081 mol

Now find out the mass of CO₂

    mass = number of moles × molar mass

    mass of CO₂ = 0.081 mol × 44.01 g/mol

    mass of CO₂ = 3.565 g

Convert mass to volume

    volume = mass / density

    V of CO₂ = 3.565 g / 1.97 g / L

    V of CO₂ = 1.81 L

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Explanation:

The 18th-century chemist Antoine Lavoisier and his wife Marie-Anne put fruit into a glass container, and then they sealed it. After weighing the fruit and container, they left it in a warm place for a few days. The fruit decayed, and there was condensation inside the glass and gas in the container. Lavoisier and his wife weighed the decayed fruit and the container again. The weight was the same. What scientific principle did the Lavoisier's experiment support?

Answer:

There will be the formation of a Common salt (NaCl) and Mg(OH)2 which is a base after the reaction between the two chemicals

Since the start of the industrial revolution more than 200 years ago, human activity has led to an increase in the atmospheric concentration of carbon dioxide (CO2). The pH of the ocean's surface waters has decreased by 0.1 pH units throughout this time.

A significant and swift transition in an economy (as in late-18th-century England) signaled by the widespread adoption of power-driven machinery or by a significant shift in the prevalent forms and applications of such machinery.

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If you compare 1 M Nacl solution, 1M glucose solution and 1M sucrose solution,1M Nacl solution is more hypertonic due to the dissociation of ions in the solution.

Of the two solutions the solution with higher osmotic pressure is called hypertonic. The extra pressure developed on the solution side due to the diffusion of solvent from its high amount through a semipermeable membrane is called the osmotic pressure.

    Osmotic pressure= icRT

     Where, i= van't hoff factor

                  c= concentration

                   R=Gas constant

                   t= temperature

The extent of dissociation or association of a solute in solution is expressed in terms of the van't hoff factor.

In 1M Nacl, as Nacl dissociates into ions in the solution its van't hoff factor becomes 2. In  1 M glucose it doesn't dissociate into ions. So van't hoff factor remains 1. And In 1 M sucrose,as  it does not dissociate into ions van't hoff factor  remains 1. So here Nacl has more osmotic pressure.

So, Nacl is hypertonic because of its dissociation into ions in the solution.

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Answer:

1. 0.48 M  2. 0.00814 M  3. 0.1276 M

Explanation:

1. Convert the 40 g to moles using the molar mass of C6H12O6:

40 g C6H12O6/180.16 g C6H12O6 = 0.222 mol

Then divide the number of moles by liters to find molarity:

0.222 mol/1.5 L = 1.48 M

2.  Convert the 1.55 g to moles using the molar mass of KBr:

1.55 g KBr/119 g KBr = 0.013025 mol

Then divide the number of moles by liters to find molarity:

0.013025 mol/1.60 L = 0.00814 M

3. Convert the 9.5 g to moles using the molar mass of NaOCl:

9.5 g NaOCl/180.16 g NaOCl = 0.1276 mol

Then you divide the number of moles by liters to find molarity:

0.1276 mol/1 L = 0.1276 M

Answer:

The chemical compound responsible for ocean acidification in the open oceans of the world is carbon(II) oxide

The chemical compound responsible for ocean acidification in the open oceans of the world is carbon(II) oxide.

Ocean acidification is the global decrease in seawater pH caused by the oceans' extensive uptake of atmospheric carbon dioxide (CO2) in the atmosphere. Ocean acidification is primarily caused by the Earth's atmosphere being overburdened with massive amounts of CO2, which is created by automobiles, industrial processes, and agricultural activities.

A third to a half of the CO2 produced into the atmosphere by human activity since the start of the Industrial Revolution, or around 1750, is being absorbed by the oceans. Scientists believe that at that time, seawater's average pH decreased between 8.19 to 8.05, representing a 30% increase in acidity. The chemical compound responsible for ocean acidification in the open oceans of the world is carbon(II) oxide.

Therefore, the chemical compound responsible for ocean acidification in the open oceans of the world is carbon(II) oxide.

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Answer:

the missing component is the option (a) electron

Explanation:

this is a nuclear decay of Th which resulted in the emission of alpha particles, equivalent to high speed electrons.

The reaction has a low activation energy and mostly of the reactants are converted into products hence the reaction tends to occur in the forward direction

2) The rate of the reverse reaction is too small compared to  the rate of the forward reaction.

The term reaction has to do with the interaction that occurs between the reactants and the products. Thus as the reactants interacts, a new substance is formed and then we have new substances and they are called  the products. Looking at the reaction as it is written in question (1), we can see that the there is a complete and total interaction that occurs as all the reactants are converted into products. Given the fact that all the reactants are converted into products, the reactions can only proceed forward.

In this reaction, we can see from the equation that the reaction is reversible. This means that the reaction is able to occur in both directions; forward and backwards. This reactions proceeds mostly forward hence the rate of reverse reaction is infinitesimal compared to the rate of the forward reaction.

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the concentration of Na after the two solutions are mixed together is 0.662 M.

1. Volume of Na2CrO4 = 132.4 ml = 0.1324 L.

2. Concentration of Na2CrO4 = 2.50 M

3. Volume of AgNO3 = 125 ml = 0.125 L

4.  Concentration of AgNO3 = 2.50 M.

Calculation :-

1. Moles of Na2CrO4 =  Molarity * Volume in L.

Moles of Na2CrO4= 2.50 * 0.1324 = 0.331 mol

2. Moles of AgNO3 = Molarity * volume in L

Moles of AgNO3 = 2.50 * 0.125 = 0.3125 mol

Solution :

the reaction of aqueous silver nitrate (AgNO3) with aqueous sodium chromate (Na2CrO4) to form solid silver chromate (Ag2CrO4) and aqueous sodium nitrate (NaNO3).

2 AgNO3(aq) + Na2CrO4 (aq) \rightarrow Ag2CrO4(s) + 2 NaNO3(aq).

Above balanced reaction shows that,

2 moles of AgNO3 reacts with 1 mol Na2CrO4.

Let,  0.3125 mol AgNO3 reacts with x mol Na2CrO4.

x = 0.3125 / 2

x = 0.1563 mol Na2CrO4.

Now,

1. Balanced reaction shows that,

2 mol AgNO3 forms 2 mol NaNO3.

So, 0.3125 mol AgNO3 forms 0.3125 mol NaNO3.

So,  moles of NaNO3 = 0.3125 mol.

As 1 molecule of NaNO3 gives 1 Na^+ ion.

0.3125 mol NaNO3 gives 0.3125 mol Na^+ ions...... A

Aslo,

2. remaining moles of Na2CrO4 = total moles - used moles

Remaining moles of Na2CrO4 = 0.331 - 0.1563 = 0.1748 mol.

1 molecule of Na2CrO4 gives 2 Na^+ ions.

So, 0.1748 moles of Na^2+ ions gives 0.3496 mol Na^+ ions............ B

3.So, total moles of Na^+ ions= 0.3496 + 0.3125......(from A and B)

Total moles of Na^+ ions = 0.662 mol.

So, concentration of Na^+ ions = 0.662 M.

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