It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much concentrated nitric acid is required to make the desired solution?

Answer:

26mL

Explanation:

NaOH+HCl= NaCl+H2O

nHCl=0.0243*2=0.0486

nNaOh=nHCl

VNaOH=0.0486/1.87=0.026l=26ml

Answer:

26.0 mL

Explanation:

Explanation:

since hydrochloric acid is an acid and calcium hydroxide is a base, we know that this is an acid base reaction. The ions will then dissociatiate and bond with one another.

Explanation:

here's the answer to your question about

The number of replaceable electrons in an atom is called its valency.

Examples

Thanks !

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Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

Answer:

1.28 M

Explanation:

Step 1: Given data

Mass of luminol (solute): 17.0 g

Volume of water: 75.0 mL (this is also the volume of solution)

Step 2: Calculate the moles corresponding to 17.0 g of luminol

The molar mass of luminol is 177.16 g/mol.

17.0 g × 1 mol/177.16 g = 0.0960 mol

Step 3: Calculate the molarity of the solution

We will use the definition of molarity

M = moles of solute / liters of solution

M = 0.0960 mol / 0.0750 L = 1.28 M

Answer:

a. 1.12 L

Explanation:

Step 1: Write the balanced equation for the photosynthesis

6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)

Step 2: Calculate the moles corresponding to 2.20 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

2.20 g × 1 mol/44.01 g = 0.0500 mol

Step 3: Calculate the moles of O₂ produced

The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol

Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP

At STP, 1 mole of O₂ occupies 22.4 L.

0.0500 mol × 22.4 L/1 mol = 1.12 L

Answer:

p'PCl3 =  6.8 torr

p'Cl2 =26.4 torr

p'PCl5 =223.4 torr

Explanation:

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is

PCl3(g) + Cl2(g) ---> PCl5(g)

Calculate the new partial pressures after equilibrium is reestablished. [in torr]

pPCl3

pCl2

pPCl5

Step 1: Data given

Partial pressure before adding chlorine gas:

Partial pressure of PCl5 = 217.0 torr

Partial pressureof PCl3 = 13.2 torr

Partial pressureof Cl2 = 13.2 torr

A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing

Step 2: The equation

PCl3(g)+Cl2(g) ⇔ PCl5(g)

Step 3: The expression of an equilibrium constant before adding chlorine gas

Kp = pPCl5 / (pPCl3 * pCl2)

Kp = 217.0 / (13.2 * 13.2)

Kp =  1.245

Step 4:  The expression of an equilibrium constant after adding chlorine gas

Partial pressure of PCl5 = 217.0 torr

Partial pressure of PCl3 = 13.2

Partial pressure of Cl2 = TO BE DETERMINED

Step 5: The total pressure of the system

Ptotal = pPCl5 + pPCl3 + pCl2

263.0 torr = 217.0 torr + 13.2 torr + pCl2

pCl2 = 263.0 - 217.0 -13.2 = 32.8 torr

Step 6: The initial pressure

The equation: PCl3(g)+Cl2(g) ⇔ PCl5(g)

pPCl3 = 13.2 torr

pCl2 = 32.8 torr

pPCl5 = 217.0 torr

Step 7: The pressure at the equilibrium

p'PCl3 = (13.2 -x) torr

p'Cl2 = (32.8 - x) torr

p'PCl5 = (217.0 + x) torr

Step 8: The equilibrium constant

'Kp =  p'PCl5 / (p'PCl3 * p'Cl2)

1.245 = (217.0+x) / ((13.2-x)(32.8-x)

x = 6.40 torr

p'PCl3 = 13.2 -6.40 = 6.8 torr

p'Cl2 = 32.8 - 6.40 =26.4 torr

p'PCl5 = 217.0 + x) 6.4 = 223.4 torr

Answer:

Option D. 17.5

Explanation:

Equiibrium is: CO + 2H₂  ⇄  CH₃OH

1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.

Initially we have 0.42 moles of CO and 0.42 moles of H₂

If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.

So in the equilibrium we may have:

0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂

Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen

Finally 0.13 moles of methanol, are found after the equilibrium reach the end.

Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²

0.13 / (0.29 . 0.16²)

Kc = 17.5

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Answer:

The order of reactivity towards electrophilic susbtitution is shown below:

a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene

b. p-cresol>p-xylene>toluene>benzene

c.Phenol>propylbenzene>benzene>benzoic acid

d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene

Explanation:

Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.

For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.

The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.

Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.

Answer:

[tex]x=54g[/tex]

Explanation:

From the question we are told that:

Content of Sterling silver:

Let x be sterling silver

Silver [tex]S=0.9x[/tex]

Copper [tex]C=0.1x[/tex]

Total  silver available [tex]T=48.3[/tex]

Generally the equation for Total amount to be made is mathematically given by

[tex]T=\frac{x*90}{100}[/tex]

[tex]x=\frac{48.3*100}{90}[/tex]

[tex]x=54g[/tex]

[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]

Balanced Equation:-

[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]

 9 is the element Florine

Florine has 9 electrons as well as the 9 protons that determine its atomic number.

The ground state configuration is the lowest energy configuration.

Answer:

The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.

0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3

0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.

Answer:

pH = 3.02

Explanation:

Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:

HOAc ⇄ H⁺ + OAc⁻.

In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.

By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02

______________________________________________________

*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.

Answer:

[tex]\lambda=3451*10^{10}m[/tex]

Explanation:

From the question we are told that:

Energy state [tex]e=3.50 eV[/tex]

Time [tex]t=2ms[/tex]

Generally the equation for energy of Photon is mathematically given by

[tex]E=e-e_0[/tex]

[tex]E=3.6*10^{-19}J[/tex]

[tex]E=5.7*10^{-19}J[/tex]

Generally the equation for Wave-length of Photon is mathematically given by

[tex]\lambda=\frac{hc}{E}[/tex]

[tex]\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}[/tex]

[tex]\lambda=3451*10^{10}m[/tex]

Answer:

Sorry but I know only what ksf stand for

Explanation:

Ksf stand for solubility product constant

Answer:

ksp stands for solubility product constant .

kf stands for molal freezing point depression constant ..

Explanation:

KSP = The solubility product constant, Ksp​, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the Ksp value it has .

KF = Kf is a constant for a given solvent. Kf is called the molal freezing point depression constant and represents how many degrees the freezing point of the solvent will change when 1.00 mole of a nonvolatile nonionizing (nondissociating) solute dissolves in one kilogram of solvent.

Answer:

23

Explanation:

Answer:

[tex]m_{H_2O}=12.9gH_2O[/tex]

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the reaction whereby butane is combusted in the presence of excess oxygen:

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

Thus, we can evidence a 2:10 mole ratio of butane to water, and thus, the stoichiometric setup to calculate the mass of produced water is:

[tex]m_{H_2O}=7.49gC_4H_{10}*\frac{1molC_4H_{10}}{52.12gC_4H_{10}} *\frac{10molH_2O}{2molC_4H_{10}}*\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=12.9gH_2O[/tex]

Regards!

Answer:

"They established the laws of planetary motion"

Explanation:

Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."

Answer:

is this question or you just asking I can't understand.

Answer:

one mole of water (6.022 x 10 23 molecules) has a mass of 18.02 g. One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.

Explanation:

• The mole (or mol) represents a certain number of objects.

• SI def.: the amount of a substance that contains the same

number of entities as there are atoms in 12 g of carbon-12.

• Exactly 12 g of carbon-12 contains 6.022 x 10 23 atoms.

• One mole of H 2O molecules

contains 6.022 x 10 23 molecules.

• 1 mole contains 6.022 x 10 23 entities (Avogadro’s number)

• One mole of NaCl contains 6.022 x 10 23 NaCl formula units.

• Use the mole quantity to count formulas by weighing them.

• Mass of a mole of particles = mass of 1 particle x 6.022 x 1023

Mass of 1 H atom: 1.008 amu x 1.661 x10-24 g/amu = 1.674 x10-24 g

Mass of 1 mole of H atoms:

1.674 x10-24g/H atom x 6.022 x1023H atoms = 1.008 g  

• The mass of an atom in amu is numerically the same

as the mass of one mole of atoms of the element in grams.

• One atom of sulfur has a mass of 32.07 amu;

one mole of S atoms has a mass of 32.07 g

Answer:

See Explanations

Explanation:

pH =-log[H₃O⁺] = -log[H⁺]

pOH = -log[OH⁻]

For weak acids [H⁺] = SqrRt(Ka·[Acid])

For weak bases [OH⁻] = SqrRt(Kb·[Base])

pH + pOH = 14

__________________________________________

A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95

CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;

[OH⁻]  = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M

=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05

=> pH = 14 - pOH = 14 - 2.05 = 11.95.

*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.

___________________________________________________

B. Given 0.18M CH₃NH₃Cl

In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴

Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.

Hydrolysis Reaction of Methylammonium Ion:

CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻

Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹                                   Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.

*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.

C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl          

Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl

In Water ...

=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl

=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻

=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻

-----------------------------------------------------------

Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)

=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹

----------------------------------------------------------

From the 0.36M CH₃NH₃⁺

=>       CH₃NH₃⁺ + H₂O  ⇄ CH₃NH₄⁺ + OH⁻

C(eq)   0.36M        ----              x             x     (<= at equilibrium after mixing)

Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)

=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M

=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process

=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻

--------------------------------------------------------

∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix

The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and

CH₃NH₃Cl and the mixture give the following approximate values;

A. The pH value of the 0.18 M CH₃NH₂ is 11.93

B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69

C. The pH value of the mixture is 10.644

A. 0.18 M CH₃NH₂

The solution is presented as follows;

CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻

Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we

have;

The number of moles of CH₃NH₂ remaining = 0.18 - x

Which gives;

[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]

[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴

Therefore;

[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]

4.167 × 10⁻⁴ × (0.18 - x) = x²

4.167 × 10⁻⁴ × (0.18 - x) - x² = 0

Which gives;

x = [OH⁻] = 8.455 × 10⁻³

pH = 14 + log[OH⁻]

Which gives;

pH = 14 + log(8.455 × 10⁻³) ≈ 11.93

B.  0.18 M CH₃NH₃Cl

The solution is presented as follows;

CH₃NH₃⁺ → CH₃NH₂ + H⁺

Let x represent the number of moles of CH₃NH₂ and H⁺ produced,

respectively, we have;

The number of moles of CH₃NH₃⁺ remaining = 0.18 - x

Which gives;

[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]

Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹

Therefore;

[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]

2.27 × 10⁻¹¹ × (0.18 - x) = x²

2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0

Which gives;

x = [H⁺] ≈ 2.02 × 10⁻⁶

pH = -log[H⁺]

Which gives;

pH = -log(2.02 × 10⁻⁶) ≈ 5.69

C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;

Based on the Henderson-Hasselbalch equation, we have;

[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]

Which gives;

[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]

Learn more about Henderson-Hasselbalch equation here:

https://brainly.com/question/13651361

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Hope that helps you!

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[tex]SilentNature[/tex]

Answer:

M of HNO₃ is 0.119M

Explanation:

A basic concept of titration is that in equivalence point:

mmoles of acid = mmoles of base

We have data from base and we only have data from volume of acid.

In a case our titration is a strong acid against a strong base.

We apply formula:

M of acid . Vol of acid = M of base . Vol of base

M of acid . 21.8 mL = 0.137M . 19 mL

M of acid = (0.137M . 19 mL) / 21.8 mL

M of acid = 0.119 M

When we neutralize all the titrant we reach the equivalence point.

At this point, pH = 7

2HNO₃  +  Ca(OH)₂ → Ca(NO₃)₂ +  2H₂O

Answer:

Low value for copper recovery

Explanation:

The percentage recovery is obtained from;

Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100

Note that the fact that some of the copper nitrate solution splashed out of the beaker means that some amount copper has been lost from the system. This loss of copper leads to a lower value of copper recovered from solution.

Answer:

Explanation:

Your mom

Answer:

F 64.6 percent of carbon may be

Answer:

True

Explanation:

The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).

Answer:

The molar mass is "185.44 g/mol".

Explanation:

According to the question,

The moles of NaOH will be:

= [tex]\frac{0.315}{1000}\times 25.72[/tex]

= [tex]8.1018\times 10^{-3} \ moles[/tex]

Number of moles of an acid will be:

= [tex]\frac{8.1018\times 10^{-3}}{3}[/tex]

= [tex]2.7006\times 10^{-3} \ moles[/tex]

We know that,

⇒ [tex]Moles = \frac{Mass}{Molar \ mass}[/tex]

hence,

Molar mass of unknown acid will be:

= [tex]\frac{Mass \ in \ g}{Moles}[/tex]

= [tex]\frac{0.5008}{2.7006\times 10^{-3}}[/tex]

= [tex]185.44 \ g/mol[/tex]