The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.20 atm and H2 is 0.15 atm?

Answer:

2KHCO

3

+H

2

SO

4

→K

2

SO

4

+2CO

2

+2H

2

O

Answer:

b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.

Explanation:

Answer:

it is propane

C3H8 it is propane

The Full name of the given compound is Propane.

Propane is a three-carbon alkane with the molecular formula C₃H₈.

It is a gas at standard temperature and pressure, but compressible to a transportable liquid.

In the figure given ;

In the given figure, there is single bond present between Carbon and Hydrogen. Hence, The Full name of the given compound is Propane.

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Answer:

F 64.6 percent of carbon may be

Answer:

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Explanation:

Given the data in the question;

wavelength of proton λ = 231 nm = 231 × 10⁻⁹ m

we determine the energy of the proton;

E = hc / λ

where h is plank constant ( 6.626 × 10⁻³⁴ JS )

and c is the speed of light ( 3 × 10⁸ m/s )

we substitute

E = [ ( 6.626 × 10⁻³⁴ JS ) × ( 3 × 10⁸ m/s ) ] / [ 231 × 10⁻⁹ m ]

E = 8.61 × 10⁻¹⁹ J

we know that, bond energy for H-I is 295 kJ/mol

so, H = 295 × 10³ J/mol

Now, energy to dissociate HI will be;

⇒ H / N

where N is the Avogadro's number ( 6.023 × 10²³ mol⁻¹ )

energy to dissociate HI = ( 295 × 10³ J/mol ) / ( 6.023 × 10²³ mol⁻¹ )

= 4.898 × 10⁻¹⁹ J

Therefore, Excess energy over dissociation will be;

⇒ ( 8.61 × 10⁻¹⁹ J ) - ( 4.898 × 10⁻¹⁹ J )

= 3.712 × 10⁻¹⁹ J

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

Answer:

by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32

Answer:

for you ta bangla de Mayo for bangla ki you would know it has changed an anyone come by with home home but is in an going forward you as as on Saturday to what our current meeting office next weekend too with we before too

Explanation:

rdirr5idj5rxu5ujx464jzm64izu6jz46ukz454ukzutzne4i6kxl6iz46irlz6ikzrkdrrdkyjfr6jftjtcyyjtfjtfyjtfygtukk6g744C hubungi advisers seg4ygs4e for na bangla na for de you will could would need some you any other than hand ideas suggestions is still there still just going forward forward with and my you he that was there can go over on there about

Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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Answer

Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)

Answer:

a) K2[Ni(CN)4]

b) Na3[Ru(NH3)2(CO3)2]

c) Pt(NH3)2Cl2

Explanation:

Coordination compounds are named in accordance with IUPAC nomenclature.

According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.

The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.

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:

o

Explanation:

it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.

Answer:

True

Explanation:

Aromatic rings undergo nucleophillic substitution reactions in the presence of a electron withdrawing group which stabilizes the Meisenheimer complex.

When the nucleophile attacks the ring carbon atom carrying the eventual leaving group. A resonance-stabilized carbanion intermediate called a Meisenheimer complex is formed.

Subsequent loss of the leaving group from the intermediate complex yields the product of the reaction.

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:

The options are incorrect.................

Explanation:

The Oxidation no. is +3

Answer:

12

Explanation:

2+8+2=12

atomic no is the No of protons

Answer:

Atomic number is 12.

Explanation:

Atomic number = electrons in filled shells + outermost electrons

= 2 + 8 +2

= 12

Answer:

the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed

Answer:

See explanation and image attached

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.

Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.

The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.

Answer:

XSO₄

Explanation:

XCl₂ is the chloride of metal X. The sum of the charges of the cation and the anion must be zero because the salt is electrically neutral. The charge of  the cation of X is:

1 × X + 2 × Cl = 0

1 × X + 2 × (-1) = 0

X = +2

X has a charge +2 and sulphate (SO₄²⁻) a charge -2. The neutral salt they form is XSO₄.

Answer:

5.0 × 10⁻⁶

Explanation:

Step 1: Write the balanced equation for the solution of chromium(III) iodate

Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

Step 2: Calculate the solubility product constant (Ksp)

To relate Ksp and the solubility (S), we will make an ICE chart.

        Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

I                               0                 0

C                             +S              +3S

E                                S               3S

The solubility product constant is:

Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶

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]

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Answer:

a

...

........

...........

Answer:

0.2M H2C6H5O7 < 0.2M H2C2O4

Explanation:

A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.

More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.

Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.

H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.

H₂C₆H₅O₇ < H₂C₂O₄

The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)

Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.

Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.

Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.

Strong acids ionizes completely to produce hydrogen ions.

Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.

In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.

H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23

H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08

Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇

For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.

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Answer:

[tex]\boxed {\boxed {\sf 0.566 \ mol \ Fe}}[/tex]

Explanation:

We are asked to convert a number of atoms to moles.  

We can convert atoms to moles using Avogadro's Number, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of iron (Fe). There are 6.022 ×10²³ atoms of iron in 1 mole of iron.  

We use dimensional analysis to convert atoms to moles. This involves setting up ratios. Use Avogadro's Number and the underlined information to make a ratio.

[tex]\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]

We are converting 3.41 × 10²³ atoms of iron to moles, so we multiply by this value.

[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]

Flip the ratio. It stays equivalent, but it allows the units of atoms of iron to cancel.

[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac{1 \ mol \ Fe} {6.022 \times 10^{23}\ atoms \ Fe}[/tex]

[tex]3.41 \times 10^{23}*\frac{1 \ mol \ Fe} {6.022 \times 10^{23}}[/tex]

[tex]\frac{3.41 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ Fe[/tex]

[tex]0.5662570575\ mol \ Fe[/tex]

The original measure ment of iron atoms ( 3.41 × 10²³ ) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 2 in the ten-thousandths place ( 0.5662570575) tells us to leave the 6 in the thousandths place.

[tex]0.566 \ mol \ Fe[/tex]

3.41 × 10²³ atoms of iron is equal to approximately 0.566 moles of iron.

Answer:

"[Temperature is a measurement of the average kinetic energy of the molecules in an object or a system. Kinetic energy is the energy that an object has because of its motion. The molecules in a substance have a range of kinetic energies because they don't all move at the same speed.]"

Answer:

Temperature is directly proportional to the average translational kinetic energy of molecules in an ideal gas

Explanation:

Answer:

We slip when we step on a banana peel because the inner side of banana peel being smooth and slippery reduces the friction between the sole of our shoe and the surface of road.

Answer:

answer is A. Endocrine system

Endocrine glands secrete hormones straight into the bloodstream. Hormones help to control many body functions, such as growth, repair and reproduction.

Answer:

A endocrine system

this is the answer

Answer:

electric charge

Explanation:

Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.

When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.

A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.

A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.