Determine whether the following statement about reaction rates is true or false. If the statement is false, select the reason why.
Increasing the temperature of a reaction system decreases the activation energy of the reaction.
A. True
B. False

Answer:

quartz

Explanation:

The correct option would be quartz.

Allotropy is a phenomenon that describes the natural existence of the same element in different forms with different physical characteristics. Allotropes are therefore different forms of the same element.

Carbon as an element has several allotropes which include diamond, graphite, graphene, amorphous carbon, and fullerenes. Quartz is a crystalline solid that is composed of silicon dioxide and not carbon.

Hence, all the options are carbon allotropes except quartz.

Answer:

Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.

Explanation:

I hope this helps bro

Answer:

The equilibrium concentration of ICl is 2.26 M

Explanation:

Chemical equilibrium is a state in which no changes are observed as time passes, despite the fact that the substances present continue to react. This is because chemical equilibrium is established when the forward and reverse reaction take place simultaneously at the same rate.

For the study of chemical equilibrium, the so-called equilibrium constant Kc is useful. Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is:

[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

For the reaction:

2 ICl (g) ⇌ I₂ (g) +Cl₂ (g)

the constant Kc is:

[tex]Kc=\frac{[I_{2} ]*[Cl_{2} ]}{[ICl]^{2} }[/tex]

Being Kc =0.110 and the concentration being the amount of moles of solute that appear dissolved in each liter of the mixture and being calculated by dividing the moles of the solute by the liters of the solution:

and replacing in the constant we get:

[tex]0.110=\frac{0.750*0.750}{[ICl]^{2} }[/tex]

Solving, you get the ICl concentration at equilibrium:

[tex][ICl]^{2} =\frac{0.750*0.750}{0.110 }[/tex]

[tex][ICl] =\sqrt{\frac{0.750*0.750}{0.110 }}[/tex]

[ICl]= 2.26 M

The equilibrium concentration of ICl is 2.26 M

Answer:

Pentan-2-ol

Explanation:

On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ([tex]H^+[/tex])  would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".

See figure 1

I hope it helps!

Answer:

The acid reacts with the conjugate base producing more weak acid.

Explanation:

A buffer solution is defined as the mixture of a weak acid and its conjugate base or a weak base with its conjugate acid.

The acetic buffer, CH₃COOH/CH₃COO⁻, is in equilibrium with water as follows:

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺

When an acid HX (Source of H₃O⁺) is added to the buffer, the reaction that occurs is:

CH₃COO⁻ + HX → CH₃COOH

In fact, this is the principle of the buffer:

An acid reacts with the conjugate base producing weak acid. And the weak acid reacts with a base producing conjugate base

If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.

We have a buffer formed by acetic acid and sodium acetate.

A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.

They can be formed in 1 of 2 ways:

Our buffer is formed by a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate).

When an acid (HX) is added, it is neutralized by the basic component of the buffer. The generic net ionic equation is:

H⁺ + CH₃COO⁻ ⇄ CH₃COOH

If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.

Learn more about buffers here: https://brainly.com/question/24188850

Answer:

a.

[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]

b.

[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]

c.

[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]

Explanation:

Hello,

In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:

a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]

[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]

b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]

[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]

c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]

[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]

Regards.

Answer:

This is due to the water moisture present in the recovered sample.

Explanation:

The total amount of material recovered isn’t meant to weigh more than the original sample. However when this happens then it means there is the presence of water moisture in the recovered sample.

The recovered samples however needs to be heated to make it dry and eliminate the water moisture through evaporation.

Answer:

0.5 mol MgCl₂

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCl → MgCl₂ + H₂

In words, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of MgCl₂ and 1 mole of H₂.

Step 2: Establish the appropriate molar ratio

The molar ratio of HCl to MgCl₂ is 2:1.

Step 3: Calculate the moles of MgCl₂ produced from 1 mole of HCl

1 mol HCl × (1 mol MgCl₂/2 mol HCl) = 0.5 mol MgCl₂

Answer:

it is 2.0, the above one is wrong

Explanation:

I did the test :

Answer:

[tex]m_{H_3PO_4}=31.5gH_3PO_4[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]P_2O_5(s)+3H_2O(l)\rightarrow 2H_3PO_4(aq)[/tex]

Thus, since the diphosphorus pentoxide to water molar ratio is 1:3 and we are given the mass of both of them, for the calculation of the maximum mass phosphoric acid that is yielded, one could first identify the limiting reactant, for which we compute the available moles of diphosphorus pentoxide (molar mass 142 g/mol):

[tex]n_{P_2O_5}=22.8gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5}=0.161molP_2O_5[/tex]

And the moles of diphosphorus pentoxide that are consumed by 13.5 g of water (molar mass 18 g/mol):

[tex]n_{P_2O_5}^{consumed}=13.5gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molP_2O_5}{3molH_2O} =0.25molP_2O_5[/tex]

Hence, since less moles of diphosphorus pentoxide are available, we sum up it is the limiting reactant, therefore, the maximum mass of phosphoric acid (molar mass 98 g/mol) is computed by considering the 1:2 molar ratio between them as follows:

[tex]m_{H_3PO_4}=0.161molP_2O_5*\frac{2molH_3PO_4}{1molP_2O_5} *\frac{98gH_3PO_4}{1molH_3PO_4} \\\\m_{H_3PO_4}=31.5gH_3PO_4[/tex]

Regards.

Answer:

-69 kJ

Explanation:

Step 1: Given data

Standard cell potential (E°cell): +0.24 V

Electrons involved (n): 3 mol

Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell

We will use the following expression.

ΔG° = -n × F × E°cell

where,

F is Faraday's constant (96,485 C/mol e⁻)

ΔG° = -n × F × E°cell

ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V

ΔG° = -69 kJ

Answer:

Cells are the basic structures of all living organisms. Cells provide structure for the body, take in nutrients from food and carry out important functions. ... These organelles carry out tasks such as making proteins?, processing chemicals and generating energy for the cell

Answer: I absolutely love this question! Biology is so interesting, so I always love to answer the curiosity of others regarding biology, such as that!

Cells are simply the basic structures of all organisms, that are living, of course! Cells provide structure for the body, and they also take in nutrients that your body needs from food and they carry out important functions. These organelles carry out tasks such as making proteins, processing chemicals, and generating energy for the cell. Isn’t that cool?

Hope this helped! <3

Answer:

It has 2

Explanation:

The significant figures are 1 and 5!

Hope this helps:)

Answer:

(D) The reaction cannot take place in a single elementary step

Explanation:

Statements are:

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I].

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution.

(D) The reaction cannot take place in a single elementary step.

The rate of the reaction is:

rate = k[CH3I][N3–].

That means rate depends of concentration of CH₃I as much as N₃⁻ concentration

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. FALSE. The reaction rate depends of N₃⁻ as much as CH₃I

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. FALSE. Reaction is second-order. Half-life is 1/K[A]₀. If initial concentration is 0.1M, to a concentration of 0.005M it takes:

1/K*0.1. If initial concentration is 0.005M it takes 1/K*0.005. That means it takes half to decrease from 0.005M to 0.0025 as it does for it to decrease from 0.01M to 0.005M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution. FALSE. Reaction rate is independent of I⁻

(D) The reaction cannot take place in a single elementary step. TRUE. As this reaction is a single-replacement reaction implies the formation  of 1 C-N bond. But also the rupture of the C-I bond is impossible to explain this kind of reaction in a single elementary step.

Answer:

There are no forces holding the gas particles together.

Explanation:

Answer:

Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.

NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?

Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.

NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?

Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.

Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?

Answer:

The binding energy = 8.64972649×10¹⁰ kJ/mole

Explanation:

Given that:

An atom of 108Te has a mass of 107.929550 amu.

In a 108 Te atom, there are 52 protons and 56 neutrons

where;

mass of proton= 1.007825 amu

mass of neutron= 1.008665 amu

Similarly; The atomic number of Te = 52

the mass of 52 protons = 52 ×  1.007825  amu

the mass of 52 protons = 52.4069 amu

the mass of 56 neutrons = 56 ×  1.008665 amu

the mass of 56 neutrons = 56.48524 amu

The total mass can now be = the mass of 52 protons + the mass of 56 neutrons

The total mass = 52.4069 amu +  56.48524 amu

The total mass = 108.89214  amu

Recall  : it is given that An atom of 108Te has a mass of 107.929550 amu.

Therefore, the mass defect will be = 108.89214  amu - 107.929550 amu

the mass defect = 0.96259amu

where 1 amu = 1.66× 10⁻²⁷ kg

Therefore, 0.96259amu = (0.96259  × 1.66× 10⁻²⁷) kg

= 1.5978994 × 10⁻²⁷kg

The binding energy = mass defect × (speed of light)²

where;

speed of light c = 2.99792 × 10⁸ m/s

The binding energy = 1.5978994 × 10⁻²⁷kg  ×  2.99792 × 10⁸ m/s

The binding energy = 1.43611597 × 10⁻¹⁰  J

The binding energy =  1.43611597 × 10⁻¹³ kJ/atom

since 1 mole = 6.023 × 10²³ atom (avogadro's constant)

Then;

The binding energy = ( 1.43611597 × 10⁻¹³ )× (6.023 × 10²³)  kJ/mole

The binding energy = 8.64972649×10¹⁰ kJ/mole

Answer:

195.5 K

Explanation:

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

2S + 3O2 + 2H2O → 2H2SO4

Next, we shall determine the number of mole in 17.55 g of H2SO4. This can be obtained as follow:

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98 g/mol

Mass of H2SO4 = 17.55 g

Mole of H2SO4 =...?

Mole = mass /Molar mass

Mole of H2SO4 = 17.55/98

Mole of H2SO4 = 0.179 mole.

Next, we shall determine the number of mole of O2 needed for the reaction. This can be obtained as illustrated below:

From the balanced equation above,

3 moles of O2 reacted to produce 2 moles of H2SO4.

Therefore, Xmol of O2 will react to produce 0.179 moleof H2SO4 i.e

Xmol of O2 = (3 x 0.179) / 2

Xmol of O2 = 0.2685 mole

Therefore, 0.2685 mole of O2 is needed for the reaction.

Finally, we shall determine the temperature of O2. This can be obtained by using the ideal gas equation as follow:

Volume (V) of O2 = 5.01 L

Pressure (P) = 0.860 atm

Number of mole (n) of O2 = 0.2685 mole

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =..?

PV =nRT

0.860 x 5.01 = 0.2685 x 0.0821 x T

Divide both side by 0.2685 x 0.0821

T = (0.860 x 5.01) /(0.2685 x 0.0821)

T = 195.5 K

Therefore, the temperature of O2 must be 195.5 K.

Answer:

Empirical formulae is NO2

Molecular Formulae is NO2

Answer:

1.84 × 10⁻³

Explanation:

Step 1: Write the balanced equation

2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

Step 2: Calculate the initial concentration of NOBr

0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:

M = 0.143 mol / 1.00 L = 0.143 M

Step 3: Make an ICE chart

         2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

I             0.143               0           0

C              -2x               +2x        +x

E          0.143-2x            2x          x

Step 4: Find the value of x

The equilibrium concentration of NOBr(g) was 0.108 M. Then,

0.143-2x = 0.108

x = 0.0175

Step 5: Calculate the concentrations at equilibrium

[NOBr] = 0.108 M

[NO] = 2x = 0.0350 M

[Br₂] = x = 0.0175 M

Step 6: Calculate the equilibrium constant (Kc)

Kc = [0.0350]² × [0.0175] / [0.108]²

Kc = 1.84 × 10⁻³

Answer:

pH of solution B is 5

Explanation:

A weak acid, HA, is in equilibrium with water as follows:

HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)

Where Ka (10^-pKa = 1x10⁻⁹) is:

Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

Where concentrations of this species are equilibrium concentrations

As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:

[HA] = 0.1M - X

[A⁻] = X

[H₃O⁺] = X

Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.

Replacing in Ka expression:

1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

1x10⁻⁹ = [X] [X] / [0.1 - X]

1x10⁻¹⁰ - 1x10⁻⁹X = X²

1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0

Solving for X:

X = -0.00001 → False solution, there is no negative concentrations.

X = 1x10⁻⁵ → Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 1x10⁻⁵M

And pH = -log[H₃O⁺]

pH = 5

Answer:

1. oxidation

2. reduction

3. oxidation

4. oxidation

Explanation:

Oxidation and Reduction in terms of hydrogen

Oxidation and Reduction in terms of Oxygen

Answer:

d. carboxyl

Explanation:

The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.

For example:

Hence, the correct option is "d. carboxyl ".

Answer:

AgBr

Explanation:

Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.

However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.

The equation for the dissolution of AgBr in cyanide is shown below;

AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)

Answer:

0.27 V

Explanation:

Given that the both half cells contain 1.0 molar solutions of their respective electrolytes.

E°Pb= -0.13 V

E°Cd = -0.40 V

Since it is a galvanic cell, the electrode having a more negative electrode potential will serve as the anode and the electrode having a less negative electrode potential will serve as the cathode.

Hence cadmium will serve as the anode and lead will serve as the cathode.

E°cell = E°cathode - E°anode

E°cell = -0.13 - (-0.40)

E°cell = 0.27 V

Answer:

The volume that a mass of 47.2 g would occupy is 0.0334 L

Explanation:

Density is the property that matter, whether solid, liquid or gas, has to compress into a given space. Density is defined as the amount of mass it has per unit volume, that is, the ratio between the mass of a body and the volume it occupies:

[tex]Density=\frac{mass}{volume}[/tex]

This indicates that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

In this case:

Replacing:

[tex]1.413 \frac{g}{mL}=\frac{47.2 g}{volume}[/tex]

Solving:

[tex]volume=\frac{47.2 g}{1.413\frac{g}{mL} }[/tex]

volume=33.40 mL

Being 1,000 mL= 1 L:

volume= 0.0334 L

The volume that a mass of 47.2 g would occupy is 0.0334 L

Answer:

A : At the end of the hour, isotope B has a greater decay constant λ than isotope A

Explanation:

Firstly, we need to understand that radioactive decay follows a first order rate law.

What this means is that we can calculate the radioactive decay constant using the following formula from the half-life

Mathematically;

[tex]t_{1/2}[/tex]  = 0.693/λ

where λ represents the radioactive decay constant.

Rearranging the equation, we can have

λ = 0.693/[tex]t_{1/2}[/tex]

Now, to have a fair level playing ground, it is best that the half-life of both isotopes are in the same unit of time(seconds)

For A, the half-life = 2.3 minutes which is same as 2.3 × 60 = 138 seconds

For B, the half-life is 24 seconds

Thus, at the end of the hour, the decay constant for isotope A will be;

λ = 0.693/138 = 0.0050 [tex]s^{-1}[/tex]

For isotope B, the decay constant will be;

λ = 0.693/24 = 0.028875  [tex]s^{-1}[/tex]

We can see that the decay constant of isotope B is higher than that of A at the end of the experiment

Answer:

The answer is o2

Explanation:

I took the test

Answer:

there is one atom of oxygen and two atoms of hydrogen

Explanation:

Answer:

In order from lowest to highest:

Methane < Ethane < Chloroethene < Methanol

i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl

Explanation:

Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.

I'm positive this explanation would suffice. Best of luck.

The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.

Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.

The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.

Learn more: https://brainly.com/question/865531

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

[tex]left\ over=0g[/tex]

Regards.