Which of the following is an example of a formula of a compound?
A. NH3
B. Nitrogen trihydride
C. Nitrogen + Hydrogen = Nitrogen trihydride
D. Ammonia​

Answer:

3.3 g of glucose, C6H12O6.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

6CO2 + 6H2O —> C6H12O6 + 6O2

Next, we shall determine the mass of CO2 that reacted and the mass of C6H12O6 produced from the balanced equation.

This is illustrated below:

Molar mass of CO2 = 12 + (2x16) = 44 g/mol

Mass of CO2 from the balanced equation = 6 x 44 = 264 g

Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180 g/mol

Mass of C6H12O6 from the balanced equation = 1 x 180 = 180 g

From the balanced equation above,

264 g of CO2 reacted to produce 180 g of C6H12O6.

Finally, we shall determine the mass of C6H12O6 produced by reacting 4.9 g of CO2 as follow:

From the balanced equation above,

264 g of CO2 reacted to produce 180 g of C6H12O6.

Therefore, 4.9 g of CO2 will react to produce = (4.9 x 180)/264 = 3.3 g of C6H12O6.

Therefore, 3.3 g of glucose, C6H12O6 were obtained from the reaction.

Answer:

0.6L

Explanation:

The formula of molarity is molSolute/litreSolution

[tex]0.968M=\frac{0.581}{LitreSolution} \\\\LitreSolution=\frac{0.581}{0.968} \\LitreSolution=0.6L[/tex].

Answer:

I have no clue at all im in 11 and dont know anything lol byeeee

Explanation:um um i am lost

Answer:

1.55 × 10²⁵ atoms of H  

Explanation:

3.21mol C₃H₈ × 8mol H × (6.022×10²³)

Explanation:

The law of conservation of mass states that mass can neither be created nor be destroyed.

Explanation in own words = this means that in this universe no one can create or destroy mass.

No physical or chemical force.

Answer:

D: 5s

Explanation:

hope this helps :)

Answer:

See attached picture.

Explanation:

Hello,

In this case, starting by 1-butanol, we can make it react with hydrogen bromide (1st step) in order to yield 1-bromobutane. Next, the formed alkyl halide is treated magnesium in the presence of an ether in order to yield butyl magnesium bromide which is a Grignard reagent (2nd step). Finally, by adding carbon dioxide, water and extra hydrogen chloride, a carbonyl group can be formed between two butyl radicals in order to form the 5-nonanone (3rd step) as shown on the attached picture.

Best regards.

Answer:

Explanation:

If sulfur gains 2 electrons then two electrons should be added to it electronic configuration.

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

.02M

CH₃COOH  = CH₃COO⁻ + H⁺

C                       xC             xC

Ka = xC . xC / C = x² C

1.8 x 10⁻⁵ = x² . .02

x² = 9 x 10⁻⁴

x = 3 x 10⁻²

= .03

concentration of H⁺ = xC = .03 . .02

= 6 x 10⁻⁴ M , volume =  45 x 10⁻³ L

moles of H⁺  = 6 X 10⁻⁴  x 45 x 10⁻³

= 270 x 10⁻⁷ moles

= 2.7 x 10⁻⁵ moles

concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L

moles of Na OH = 2 X 10⁻²  x 35 x 10⁻³

= 70 x 10⁻⁵ moles

=  

NaOH is a strong base so it will dissociate fully .

there will be neutralisation reaction between the two .

Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles

= 67.3 x 10⁻⁵ moles of NaOH

Total volume = 45 + 35 = 80 x 10⁻³

concentration of NaOH after neutralisation.= 67.3  x 10⁻⁵ / 80 x 10⁻³ moles / L

= 8.4125  x 10⁻³ moles / L

OH⁻ = 8.4125  x 10⁻³

H⁺ = 10⁻¹⁴ / 8.4125  x 10⁻³

= 1.1887 x 10⁻¹²

pH = - log (  1.1887 x 10⁻¹² )

= 12 - log 1.1887

= 12 - .075

= 11.925 .

Answer:

The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.  

Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.  

Considering the definition of ionization energy,

Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.

You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.

In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.

In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.

Taking into account the above, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.  

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Answer:

0.312g

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules

1mole of sulphur = 32g

If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules

Then, Xg of sulphur will contain 5.87x10^21 molecules i.e

Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g

just saying this is not my work, thank Eduard22sly

he answered it on a different page

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Answer:

Covalent bond is a chemical bond is a strong attraction between two or more atoms.

Answer:

Ionic

Explanation:

An ionic chemical bond is a strong attraction between two or more atoms. Two atoms are more strongly attracted by ionic bonds.

Answer:

The mass of NaHCO3 required is 235.22 g

Explanation:

*******

Continuation of Question:

2NaHCO3(s) + H2SO4(aq)  →  Na2SO4(aq) + 2CO2(g) + 2H2O(l)

You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?

********\

The question requires us to calculate the mass of NaHCO3  to neutralize the acid.

From the balanced chemical equation;

1 mol of H2SO4 requires 2 mol of NaHCO3

1.4 would require x?

Upon solving for x we have;

x = 1.4 * 2 = 2.8 mol of NaHCO3

The relationship between mass and number of moles is given as;

Mass = Number of moles * Molar mass

Mass = 2.8 mol * 84.007 g/mol

Mass =  235.22 g

Answer:

The process of slowly adding one solution to another until the reaction between the two is complete.

Explanation:

When you perform a titration, you are slowly adding one solution of a known concentration called a titrant to a known volume of another solution of an unknown concentration until the reaction reaches neutralization, in which the reaction is no longer taking place. This is often indicated by a color change.

Hope that helps.

Answer:

We will use an acid-base indicator to see changes in colour depending on the pH  

Explanation:

The pH changes during a titration, so you could use an acid-base indicator to follow the changes in pH.

A is wrong. An acid-base titration does not usually form a solid, and it would be impractical to isolate a solid with a funnel.

B is wrong. There are no changes in mass.

C is wrong. Any changes in temperature would be too small to measure precisely with an ordinary thermometer.

The best way to monitor the progress of a neutralization reaction such as acid-base titration: D. Use an acid-base indicator to observe the changes in color depending on the pH.

The chemical reaction that occurs when you mix an acid and a base together is referred to as neutralization reaction.

In a neutralization reaction, what is formed is salt and water.

Acid-base titration is a neutralization method.

During acid-base titration, the neutralization reaction that occurs is usually monitored by observing the pH changes that occurs.

Change in pH is an indicator that there is progress in the neutralization reaction.

An acid-base indicator, can be used to detect the changes that occur via the pH changes in relation to the color change.

Therefore, the best way to monitor the progress of a neutralization reaction such as acid-base titration: D. Use an acid-base indicator to observe the changes in color depending on the pH.

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Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)

N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..

~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

J. [,v 1+3] ~

I

4,:i.~ = 0,05

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:

[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]

At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:

[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]

We can calculate the concentrations of all the species at equilibrium using an ICE chart.

       2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)

I          1.00              0          0

C         -2x              +3x        +x

E       1.00-2x          3x          x

Since the concentration of ammonia at equilibrium is 0.200 M,

[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]

The concentrations of all the species at equilibrium are:

[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]

The concentration equilibrium constant (Kc) is:

[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

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Answer:

a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)

workdone (w) = -8442.6 J  ≈ -8.443 KJ

heat transferred (q) of the ideal gas = - w

q = 8.443 KJ

b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0

the workdone(w) in the ideal gas= - 4567.5 J  ≈ - 4.57 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Explanation:

given

mole of an ideal gas(n) = 2.5 mol

Temperature (T) = 20°C

= (20°C + 273) K  = 293 K

Initial pressure of the ideal gas(P₁) = 20 atm

Final pressure of the ideal gas(P₂) = 5 atm.

2) (a)for adiabatic reversible process,

note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.

Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]

= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]

= 6090.01 J × [-1.3863]

= -8442.6 J  ≈ -8.443 KJ

So, the work done (w) of ideal gas = -8.443 KJ

For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-8.443 KJ)

q = 8.443 KJ

heat transfer (q) of the ideal gas = 8.443 KJ

(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.

Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )

= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]

= - 6090.01 J × 0.75

= - 4567.5 J  ≈ - 4.57 KJ

∴work done(w) of an ideal gas = - 4.57 KJ

For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-4.5675 KJ)

q = 4.5675 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Answer:

pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

Explanation:

pKa acetic acid, CH3COOH = 4.7

It is possible to determine pH of a buffer using H-H equation:

pH = pka + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH3COONa] / [CH3COOH]

As you want a pH 4.7 buffer:

4.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = [CH3COONa] / [CH3COOH]

That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:

For pH 5.7:

5.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = log [CH3COONa] / [CH3COOH]

10 = [CH3COONa] / [CH3COOH] (1)

That means you need 10 times [CH3COONa] over [CH3COOH]

And as you know:

10mL=  [CH3COONa] + [CH3COOH] (2)

Replacing (1) in (2):

10 = 10mL + [CH3COOH] / [CH3COOH]

10[CH3COOH] = 10mL + [CH3COOH]

11[CH3COOH] = 10mL

[CH3COOH] = 0.91mL

And [CH3COONa] = 10mL - 0.91mL =

[CH3COONa] = 9.09mL

That is:

The volumes according to the pH are as follows:

(i) 5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa for pH 4.7

(ii) 0.91mL of 0.10 M CH₃COOH and 9.09mL 0.10 M CH₃COONa pH 5.7

Given that pKa of acetic acid, CH₃COOH = 4.7

The pH of a buffer using the H-H equation is given by:

pH = pKa + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH₃COONa] / [ CH₃COOH]

4.7 = 4.7 + log [CH₃COONa] / [ CH₃COOH]

0 = log [CH₃COONa] / [ CH₃COOH]

takin antilog on both sides of the equation we get:

1 = [CH₃COONa] / [CH₃COOH]

It implies that the same amount of both species is needed to make the pH 4.7 buffer.

So,

5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa makes a buffer of pH 4.7

Similarly:

5.7 = 4.7 + log [CH₃COONa] / [CH₃COOH]

1 = log [CH₃COONa] / [CH₃COOH]

takin antilog on both sides of the equation we get:

10 = [CH₃COONa] / [CH₃COOH]

10[CH₃COOH] = [CH₃COONa]

It implies that we need 10 times [CH₃COONa] as much of [CH₃COOH]

We have to prepare 10 mL of buffer, so:

10mL=  [CH₃COONa] + [CH₃COOH]

10mL = 11[CH₃COOH]

[CH₃COOH] = 0.91mL

So, [CH₃COONa] = 10mL - 0.91mL

[CH₃COONa] = 9.09mL

Therefore,

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa is required to make a buffer of pH 5.7

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Answer:

Option A

Explanation:

Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.

Answer:

A. Silicon (Si)

Explanation:

Silicon (Si) is primarily used as a semiconductor material to make electronic chips.

Answer:

Strontium is smaller

Strontium has the higher ionization energy

Strontium has more valence electrons

Explanation:

It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table

While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)

Since they are in the same period, periodic trends would be useful in evaluating their properties

In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size

Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius

Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has

In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case

Answer:

15.3 %

Explanation:

Step 1: Given data

Step 2: Calculate the mass of oxygen (mO)

The mass of the sample is equal to the sum of the masses of all the elements.

ms = mC + mH + mN + mO

mO = ms - mC - mH - mN

mO = 230 g - 136.6 g - 26.4 g - 31.8 g

mO = 35.2 g

Step 3: Calculate the mass percent of oxygen

%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %

Answer: B. The period number tells which is the highest energy level occupied by the electrons

Explanation:

Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.

hence, the correct option is B. The period number tells which is the highest energy level occupied  by the electrons.

The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.

The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.

When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.

Therefore, option B , second option is correct.

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Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

Answer:

pH = 3.56

Explanation:

The pH of the buffer producing from the mixture of formic acid and formate ion can be found using H-H equation:

pH = pka + log [A⁻] / [HA]

pH = 3.74 + log [Formate] / [formic acid]

Where [] represents molar concentrations -or moles- of formate and formic acid in the solution.

Replacing knowing moles of formic acid = 0.0600 and moles formate = 0.0400:

pH = 3.74 + log [Formate] / [formic acid]

pH = 3.74 + log [0.0400] / [0.0600]

Answer:

a. 2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)

b. 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)

c. Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)

Explanation:

A reaction is balanced when you have the same amount of atoms in reactants and products.

In the reactions:

(a) Cu₂O(s) + C(s) → Cu(s) + CO₂(g)

As a general rule, you first balance oxygen and hydrogen. In products you have 2 oxygens, then:

2 Cu₂O(s) + C(s) → Cu(s) + CO₂(g)

Carbon is balanced yet. Thus, you need just to balance Cu:

2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)

(b) Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)

Balancing oxygen:

2Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)

Sulfur is balanced yet. Now you just need to balance Cu:

2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)

(c) Cu₂CO₃(OH)₂(s) → CuO(s) + CO₂(g) + H₂O(g)

This reaction is different because the reactant is a chemical with a lot of atoms. we will first balance Cu:

Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)

Balancing copper, oxygen, hydrogen and carbon are balanced:

Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles[/tex]

[tex]\text{Moles of} CO=\frac{260}{28}=9.3moles[/tex]

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]

According to stoichiometry :

1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex]  of [tex]CO[/tex]

Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.

As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]

Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

[tex]\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%[/tex]

Therefore, the percent yield is, 91.8%

Answer:

Molarity of the sulfuric acid solution is 4.61M

Explanation:

The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:

2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻

That means, 2 moles of base react with 1 mole of sulfuric acid.

If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:

0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄

As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:

0.200 moles H₂SO₄ / 0.0434L = 4.61M

Explanation:

At 593K a particular decomposition’s rate constant had a value of 5.21×10−4 and at 673K the same reaction’s rate constant was 7.42×10−3. It was noticed that when the reactant’s initial concentration was 0.2264 M (with a 593K reaction temperature), the initial reaction rate was identical to the initial rate when the decomposition was run at 673K with an initial reactant concentration of 0.05999 M. Recall that rate laws have the form rate = k [A]x and, showing work, determine the order of the decomposition reaction.

Answer:

D phenolphthalein,pH range 8.2-10.0

Answer:

The answer is option D.

ethanol

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