Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.

a. True
b. False

Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.

Explanation:

A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.  

Answer:

2.772 seconds

Explanation:

Given that;

t1/2 = 0.693/k

Where;

t1/2 = half life of the reaction

k= rate constant

Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant

t1/2 = 0.693/2.5 x 10-1 s-1

t1/2= 2.772 seconds

Answer:

density of second liquid = 650 kg/m³

Explanation:

Given that:

The volume of the plastic block submerged inside the water  = 0.5 V

The force on the plastic block  = [tex]\rho_1V_1g[/tex]

[tex]= 0.5p_1 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

W [tex]= 0.5p_1 V_g[/tex]

[tex]\rho Vg = 0.5p_1 V_g[/tex]

[tex]\rho = 0.5 \rho _1[/tex]

where;

water density [tex]\rho _1[/tex] = 1000

[tex]\rho = 0.5 (1000)[/tex]

[tex]\rho = 500 kg/m^3[/tex]

In the second liquid, the volume of plastic block in the water = (100-23)%

= 77% = 0.7 V

The force on the plastic block is:

[tex]= 0.77p_2 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

[tex]W = 0.77p_2 V_g[/tex]

[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]

[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]

Answer:

6.214 degrees-mL/gdm

Explanation:

The specific rotation α' = α/LC where α = observed rotation, L = length of tube and C = concentration of solution.

Given that α = 1.74, L = length of cell = 50 mm = 0.50 dm and C = m/V where m = mass of glyceraldehyde = 5.60 g and V = volume = 10 ml

So, C = m/V = 5.60 g/10 ml = 0.560 g/ml

Since α' = α/LC

substituting the values of the variables into the equation, we have

α' = α/LC

α' = 1.74/(0.50 dm × 0.560 g/ml)

α' = 1.74/(0.28 gdm/l)

α' = 0.006214 °mL/gdm

α' = 6.214 °mL/gdm

α' = 6.214 degrees-mL/gdm

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Answer:

[tex]V_2=482.5mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law due to the fact that we are dealing with variable volume, temperature and pressure:

[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]

In such a way, we solve for the final volume, V2, considering that the initial volume, V1, is 500 mL, the initial temperature, T1, is 273 K (STP), the initial pressure, P1, is 1 atm (STP) and the final temperature, T2, is 325 K and the final pressure, P2, is 125 kPa (1.23 atm):

[tex]V_2=\frac{P_1V_1T_2}{T_1P_2} \\\\V_2=\frac{(1atm)(500mL)(325K)}{(273K)(1.23atm)} \\\\V_2=482.5mL[/tex]

Regards!

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

Answer:

+5

Explanation:

Answer:

three electron shells

14 electrons

14 neutrons

Explanation:

Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.

Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.

The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.

Answer:

An unsaturated solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly realize we need to calculate the grams of vanillin in 0.070 moles by using its molar mass of 152.15 g/mol:

[tex]m=0.070mol*\frac{152.15 g}{1mol} =10.65g[/tex]

Thus, since the solubility is 10.65 g per 1 L of solution, we can notice 5.00 g will complete dissolve and produce an unsaturated solution.

Best regards!

Answer:

The Hazmat products warnings or labels allowed in fc include:

1. Fully Regulated Aerosol Placard

2. Fully Regulated Flammable Solid Placard  

3. Fully Regulated Flammable  

4. Lithium-Ion/Metal Battery label

Explanation:

Hazmat products (including explosives, flammable liquids and solids, and gases, etc.) are classified as dangerous substances and materials that pose a risk to people during their storage, handling, or transportation.  The requirement for this Hazmat classification is to show that the identified products require diligence, carefulness, and alertness in handling, transporting, and storing them.  The reason for this is that mishaps can occur.  Some of them can also cause fire outbreaks.

Answer:

pH= 1.9 then [tex]H_{3} PO_{4}[/tex]

pH = 5.0 , [tex]CH_{3} COOH[/tex]

pH = 3.9 , HCOOH

As we know range left [tex]pH= pKa+/- 1[/tex]

Answer:

925.66 J/K

Explanation:

Applying,

Q = CΔt............. Equation 1

Where Q = amount of heat, C = heat capacity of the calorimeter, Δt = rise in temperature.

make C the subject of the equation

C = Q/Δt.............. Equation 2

From the question,

Given: Q = 3.86 kJ = 3860 J, Δt = 4.17K

Substitute into equation 2

C = 3860/4.17

C = 925.66 J/K

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

Answer:

Covalent bonding occurs when pairs of electrons are shared by atoms.

Explanation:

Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability.

Answer:

are in the same orbital

Explanation:

Answer:

are in the same orbit

Explanation:

Answer:

From least polar covalent to most polar covalent;

S-I< Br-Cl < N-H< Te-O

From most ionic to least ionic

Cs-F> Sr-Cl> Li- N> Al-O

Explanation:

Electro negativity refers to the ability of an atom in a bond to attract the shared electrons of the bond towards itself.

Electro negativity difference between two atoms is a key player in the nature of bond that exists between any two atoms. A large difference in electron negativity leads to an ionic bond while an intermediate difference in electro negativity leads to a polar covalent bond.

Based on electro negativity differences, the bonds in the answer have been arranged in order of increasing polar covalent nature or decreasing ionic nature.

````````````````````

Answer:

CH3OH < FCH2OH < F2CHOH < F3COH

Explanation:

Let us recall that, for a carboxylic acid, the dissociation of the acid yields;

RCOOH ⇄RCOO^- + H^+

The ease of dissociation and release of the hydrogen ion depends on the nature of the group designated R.

When R is is a highly electronegative element, the -I inductive effect causes the hydrogen to become less tightly held by the C-Cl bond.

As the number of electron withdrawing substituents increaseses, the acid ionizes much more and becomes stronger.CH3OH < FCH2OH < F2CHOH < F3COH

Hence, the order of decreasing acid strength is;

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

Answer:

0.17 g

Explanation:

Since the volume of gas collected is 132 mL, we need to find the number of moles of gas present in 132 mL.

So, number of moles, n = volume of gas, v/molar volume, V

n = v/V where v = 132 mL = 0.132 L and V = 22.4 L

So, substituting the values of the variables into the equation, we have

n = v/V

n = 0.132 L/22.4 L

n = 0.005893 mol

We then need to calculate the molar mass of CO, M = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol

Also, number of moles of gas, n = m/M where m = mass of CO and M = molar mass of CO

m = nM

m = 0.005893 mol × 28 g/mol

m = 0.165004 g

m ≅ 0.17 g to 2 significant digits

Answer:

No correct answer listed. See explanation for defense.

Explanation:

Given

C: 40.92% => 40.92g/100wt => (40.92/12)moles C = 3.41 moles O

H:   4.58% =>   4.58g/100wt =>      (4.58/1)moles H = 4.58 moles H

O: 54.50% => 54.5g/100wt =>     (54.5/16)moles O = 3.41 moles O

Empirical ratio => C : H: O => (3.41/3.41) : (4.58/3.41) : (3.41/3.41) => 1 : 1.34 : 1

=> C : H : O => 3(1 : 1.34 : 1) => 3 : 4 : 3 => Empirical Formula C₃H₄O₃

Molecular Weight = Empirical Formula Wt x N

         176.12          =           88  x  N

N = whole number multiple of empirical formula = 176.12/88 = 2

∴ Molecular Formula => (C₃H₄O₃)₂ => C₆H₈O₆

Note => Only ionic compounds (salts) have subscripts reduced to lowest whole number ratios. Molecular compounds as C₆H₈O₆ are not reduced to lowest whole number ratios. Therefore, there is no correct answer in the answer choice list for the 'Molecular Formula'. Doc :-)

Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons

#BRAINLEST LOVE❣️

Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.

Explanation:

Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:

--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.

--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens

--> The ability of carbon atoms to form single, DOUBLE or triple bonds.

The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.

Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.

endothermic absorbs heat

exothermic gives heat

a. endothermic

b. exothermic

c. endothermic

d. exothermic

a. Ice melting - endothermic

b. Water condensing on the surface - exothermic

c. Baking a cake - endothermic

d. The chemical reaction inside an instant cold pack - endothermic

e. A car using gasoline - exothermic

An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.

While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.

In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.

The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.

Learn more about the exothermic process, here:

brainly.com/question/12321421

#SPJ5

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

Answer:

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

Explanation:

An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.

The full equation is;

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

So, two electrons were lost in the process.