How are oxidation-reduction reactions related to how you use energy?

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

Answer:

False

Explanation:

Answer:

False

Explanation:

Answer:

d. End product is that product with a ketone and carboxylic acid.

Explanation:

[tex]{ \sf{NaBH_{4} : }}[/tex]

Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.

[tex]{ \sf{H _{2} O \: and \: H {}^{ + } }}[/tex]

Then acidified water is an oxidising mixture which reverses the reduction reaction.

Explanation:

Option D is your answer

Hope it helps

Answer:

1090 Torr

Explanation:

Step 1: Given data

Step 2: Convert pO₂ to Torr

we will use the conversion factor 1 atm = 760 Torr.

1.0 atm × 760 Torr/1 atm = 760 Torr

Step 3: Calculate the total pressure of the mixture (P)

The total pressure of the mixture is the sum of the partial pressures of the gases.

P = 330 Torr + 760 Torr = 1090 Torr

Answer:

Hence the correct option is the last option that is tends to gain electrons in chemical reactions to become anions.

Explanation:

Metals tend to donate electrons in chemical reactions to become cations.

Answer: The mass of copper (II) nitrate produced is 105.04 g.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of copper = 35.5 g

Molar mass of copper = 63.5 g/mol

Plugging values in equation 1:

[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]

The given chemical equation follows:

[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]

By the stoichiometry of the reaction:

If 3 moles of copper produces 3 moles of copper (II) nitrate

So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate

Molar mass of copper (II) nitrate = 187.56 g/mol

Plugging values in equation 1:

[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]

Hence, the mass of copper (II) nitrate produced is 105.04 g.

Answer:

The hydrogen must be compressed to 1333.13302 kPa.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

Replacing:

P* 10.5 L= 5.75 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 293 K

Solving:

[tex]P=\frac{5.75 moles* 0.082 \frac{atm*L}{mol*K} * 293 K}{10.5 L}[/tex]

P= 13.157 atm

If 1 atm is equal to 101.325 kPa, then 13.157 atm is equal to 1333.13302 kPa.

The hydrogen must be compressed to 1333.13302 kPa.

Answer:

Ca 2+ <K  +  <Ar<Cl  −  <S  2−

Explanation:

Ar,K  + ,Cl  − ,S  2− ,Ca  2+

 have the same number of electrons. Their radii would be different because of their different nuclear charges. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size. Hence the correct order will be Ca  

2+  <K +  <Ar<Cl  −  <S  2−

Answer:

[tex]E_2=999984KJ/mole[/tex]

Explanation:

From the question we are told that:

Factor [tex]dK=500[/tex]

Temperature [tex]T=37 C=310k[/tex]

Activation energy [tex]E=10^6kJ/mol[/tex]

Generally the Arhenius equation is mathematically given by

[tex]ln \frac{K_2}{K_1}=\frac{ E_1-E_2}{RT}[/tex]

Where

[tex]\frac{K_2}{K_1}=500[/tex]

[tex]ln 500=\frac{ 10^6-10^3-E_2}{8.314*310}[/tex]

[tex]E_2=999984KJ/mole[/tex]

The activation energy of the new reaction is 105.99 kJ/mol.

Using the Arrhenius equation;

ln(k2/k1) = -Ea2/RT2 +  Ea1/RT1

Now, from the information in the question;

k2/k1 = 500

Ea = ?

R = 8.314 JKmol-1

T2 = 37oC + 273 = 310 K

T1 = 37oC + 273 = 310 K

Substituting values;

ln (500) =- Ea2 + Ea1

6.2 = -Ea2 + 106 × 10^3 J

Ea = 106 × 10^3 J - 6.2

Ea = 105.99 × 10^3 J  or 105.99 kJ/mol

Learn more about activation energy: https://brainly.com/question/11334504

Answer:

See detailed explanation.

Explanation:

Hello there!

In this case, for the given scenario, we will proceed as follows:

1. Here, we infer that the products are iron (III) hydroxide (precipitate) and calcium chloride:

[tex]3Ca(OH)_2+2FeCl_3\rightarrow 3CaCl_2+2Fe(OH)_3[/tex]

2. In this step we firstly calculate the moles of both reactants, by using their molar masses 74.093 and 162.2 g/mol respectively:

[tex]14.0gCa(OH)_2*\frac{1molCa(OH)_2}{74.093gCa(OH)_2}=0.189molCa(OH)_2 \\\\17.0gFeCl_3*\frac{1molFeCl_3}{162.2gFeCl_3}=0.105molFeCl_3[/tex]

Now, we calculate the moles of calcium hydroxide consumed by 0.105 moles of iron (III) chloride by using the 3:2 mole ratio between them:

[tex]0.105molFeCl_3*\frac{3molCa(OH)_2}{2molFeCl_3} =0.157molCa(OH)_2[/tex]

Thus, we infer that calcium hydroxide is in excess as 0.189 moles are available for it but just 0.157 moles react and therefore, iron (III) chloride is the limiting reactant.

3. Here, we use the moles of iron (III) chloride we've just computed, the 2:2 mole ratio with iron (III) hydroxide and its molar mass (106.867 g/mol) as shown below:

[tex]0.105molFeCl_3*\frac{2molFe(OH)_3}{2molFeCl_3} *\frac{106.867gFe(OH)_3}{1molFe(OH)_3} \\\\=11.2gFe(OH)_3[/tex]

Regards!

Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of [tex]P_4[/tex] = 62.0 g

Molar mass of [tex]P_4[/tex] = 124 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]

Given mass of [tex]H_2[/tex] = 4.00 g

Molar mass of [tex]H_2[/tex] = 2 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]

The chemical equation follows:

[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]

So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]

As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]

So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]

We know, molar mass of [tex]PH_3[/tex] = 34 g/mol

Putting values in equation 1, we get:

[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]

Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g

Explanation:

The balanced chemical equation of the reaction is:

[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]

From the balanced chemical equation,

1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.

The molar mass of propane is 44.1 g/mol.

One mole of any gas at STP occupies --- 22.4 L.

Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.

Answer:

Thus, 67.2 L of CO2 is formed at STP.

Answer:

The final volume of the sample of gas is 36.287 liters.

Explanation:

Let suppose that sample of gas is a closed system, that is, a system with no mass interactions with surroundings, and gas is represented by the equation of state for ideal gases, which is described below:

[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)

Where:

[tex]P[/tex] - Pressure, in atmospheres.

[tex]V[/tex] - Volume, in liters.

[tex]n[/tex] - Molar quantity, in moles.

[tex]T[/tex] - Temperature, in Kelvin.

[tex]R_{u}[/tex] - Ideal gas constant, in atmosphere-liters per mole-Kelvin.

As we know that sample of gas experiments an isobaric process, we can determine the final volume by the following relationship:

[tex]\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}[/tex] (2)

Where:

[tex]V_{1}[/tex] - Initial volume, in liters.

[tex]V_{2}[/tex] - Final volume, in liters.

[tex]T_{1}[/tex] - Initial temperature, in Kelvin.

[tex]T_{2}[/tex] - Final temperature, in Kelvin.

If we know that [tex]V_{1} = 33\,L[/tex], [tex]T_{1} = 291.15\,K[/tex] and [tex]T_{2} = 320.15\,K[/tex], then the final volume of the gas is:

[tex]V_{2} = V_{1}\cdot \left(\frac{T_{2}}{T_{1}} \right)[/tex]

[tex]V_{2} = 33\,L \times \frac{320.15\,K}{291.15\,K}[/tex]

[tex]V_{2} = 36.287\,L[/tex]

The final volume of the sample of gas is 36.287 liters.

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

Answer:

K+ClKCl

Explanation:

because the reaction is between metal Potassium and Non-metal Chlorine

Answer:

Explanation:

a) It is a redox reaction because KCl is an ionic compounds with K having a + charge and Cl having a - charge. Originally, both have an oxidation state of 0 and not K has 1+ and Cl has 1-. Therefore, one species was oxidized and one was reduced which is indicative of a redox reactions.

b)

2K + Cl2 => 2KCl

Answer:

the answer should be B because elements do not tranform into other elements in a chemical reaction

am I right please?

Answer:

+3

Explanation:

u see sum of oxidation number in all situations have to be 0

ClO2 =-1

so Fe is +3

Answer:

1.37L

Explanation:

V2=v1×T2

______

T1

Answer:

Conversion of 2-methyl-2-butene into a secondary alkyl halide - ROOR, HBr

Conversion of 2-methyl-2-butene into a tertiary alkyl halide - HBr

Explanation:

The addition of HBr to 2-methyl-2-butene occurs in accordance to Markovnikov rule in the absence of peroxide.

According to Markovnikov rule; ''the negative part of the addendum is attached to the carbon atom bearing the least number of hydrogen atoms.'' Following the Markovnikov rule, the tertiary alkyl halide is obtained.

In the presence of peroxide, this rule is not followed and the reaction proceeds in an anti-Markovnikov way to yield a secondary alkyl halide.

Answer:

it increases and is perpendicular to the motion of the wave.

Answer:  hello the complete question is attached below

Visibility of molecular ion = m/z value of 77

Explanation:

For The molecular ion to be visible, it has to be at an m/z value of 77 and this is because molecular ions will have an m/z ratio =  molecular mass of given molecule in most cases but not always in all cases.

And the visibility is possible after the removal of CH₃ ion.

ii) Evidence in the mass spectrum that suggests peak at m/z = 77

attached below

the effects are: temperature rises, water shortages, and increased fire threats

Answer:

7 mol CuO

0.5 mol O₂

Explanation:

Step 1: Write the balanced equation

2 Cu + O₂ ⇒ 2 CuO

Step 2: Identify the limiting reactant

The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.

The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.

Since TMR > EMR, Cu is the limiting reactant

Step 3: Calculate the amount of CuO produced

7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO

Step 4: Calculate the excess of O₂ that remains

The amount of O₂ that reacts is:

7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂

The excess of O₂ that remains is:

4 mol - 3.5 mol = 0.5 mol

Answer:

Sodium dihydrogenphosphate = NaH₂PO₄

Sodium monohydrogenphosphate = Na₂HPO₄

Explanation:

A buffer solution is a solution is a solution that resists changes to its oH when a little quantity of strong acid or strong base is added to it.

They are solutions of weak acids or weak bases and their salts known as conjugate base or conjugate acids respectively for the weak acids and weak bases.

For example, a solution of the weak acid ethanoic acid and its salt or conjugate base, sodium ethanoate serves as a buffer solution.

In biochemical experiments, where the pH of the reaction medium is kept as constant and as close as possible to that of the internal environment, buffer solutions are widely used. One of the commonly used buffers is the phosphate buffer. The phosphate buffer consists of the acid salts sodium dihydrogenphosphate and sodium monohydrogenphosphate. Sodium dihydrogenphosphate serves as the weak acid while sodium monohydrogenphosphate serves as the conjugate base.

The formulas of these two compounds are given below:

Sodium dihydrogenphosphate = NaH₂PO₄

Sodium monohydrogenphosphate = Na₂HPO₄

Answer:

D. Propanol

Explanation:

C3H7OH the presence of alcohol functional group makes it propanol

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:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Answer:

Alphabet C :NADH is oxidized,CO2 is reduced

Answer:

solid to gaseous or gaseous to solid

Explanation:

Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. sublimation is most often used to describe the process of snow and ice changing into water vapor in the air without first melting into water.