How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)

Answer:

Half life = 13.197 hour

Explanation:

Given:

Old amount (A₀) = 3.2

New amount (A) = 0.4

Radiation decay time (t) = 39.6 hour

Half life = T(1/2)

Find:

Half life = T(1/2) = T

Computation:

A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]

[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2

-[27.4428 / T] = In (0.125)

-[27.4428 / T] = -2.0794

[27.4428 / T] = 2.0794

T = 13.197

Half life = 13.197 hour

Answer:

176.8 g of ammonia, NH3.

Explanation:

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

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6 g

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34 g.

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.

This can be obtained as follow:

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.

Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.

Answer:

B) geminal diol

Explanation:

Hello,

In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.

Regards.

Answer:

a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]

b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]

c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]

Explanation:

Hello,

In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:

a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]

b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]

c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]

Best regards.

Answer:

Molecular mass of HNO₃ = 63.015 g/mol

Explanation:

Note: The given question is incomplete , the given substance is nitric acid

Given:

Nitric acid (HNO₃)

Find:

Molecular mass

Computation:

Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)

Molecular mass of HNO₃ = 1.008 + 14.007 + 48

Molecular mass of HNO₃ = 63.015 g/mol

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:

All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.

Answer:

C) formaldehyde, H2C=O.

Explanation:

Hello,

In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).

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Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.

Answer:

pH = 8.72

Explanation:

This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

As this compound acts like a base, we propose this equilibrium:

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰

Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M

So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

We can avoid the quadractic equation because Kb is so small

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH = 8.72

The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Since the following equation should be used.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

Now

(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

So,

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

Now

Kw = Ka. Kb

Kb = Kw/Ka

And,

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵

= 5.55×10⁻¹⁰

Now

[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

Now

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH

= 8.72

Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

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Answer:

The correct option is (c)

Answer:

the law of conservation of mass

To record the melting point of tetradecanol, it is important the temperature is increase by little degrees after each trial.

The thermometer attached to the capillary tube was immersed in water without thermometer.

Usually heating curve describes the temperature change for a sample which passes through different states of matter.

The temperature than decreases and the solution in beaker cools down. To determine the melting point, it is important to keep the temperature consistent.

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Answer:

Answer:

a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

b. 18.17% of Fe in the sample

Explanation:

a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:

Fe²⁺ → Fe³⁺ + 1e⁻

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

5 times the iron and suming the manganese reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O

b. Moles of permanganate in the titration are:

0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻

Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:

1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =

4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:

4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =

Percent mass is:

0.2474g Fe / 1.362g sample ₓ 100 =

The mass percent of iron in the sample is 22.6%.

The net ionic equation of the reaction is;

5Fe^2+(aq) + 8H^+(aq) + MnO4^-   -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)

Number of moles of MnO4^-  = 39.42/1000 L × 0.0281 M = 0.0011 moles

If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-

x moles of Fe^2+ reacts with 0.0011 moles

x =  5 moles × 0.0011 moles/1 mole

x = 0.0055 moles

Mass of Fe^2+ =  0.0055 moles × 56 g/mol = 0.308 g

Mass percent of iron = 0.308 g/ 1.362 g × 100/1

= 22.6%

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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:

Carbon-14

Explanation:

It is because it have 6 protons and 8 neutrons which are added together to form an atomic mass of 14.

The mass number of an isotope is the sum of the number of protons and number of neutrons hence the correct answer is carbon-14.

Isotopes are atoms of the same element that has the same number of protons but different number of neutrons.

The atomic number is the same as the number of protons in the atom. Since there are six protons in the atom, the element must be carbon.

There are eight neutrons meaning that the mass  number is 8 + 6 = 14

Therefore, the isotope is carbon-14.

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Answer:

False because it is false

Answer: 32 electrons

Explanation:

The name of a compound is determined by IUPAC nomenclature

The International Union of Pure and Applied Chemistry have designated a universal rule for the nomenclature of organic compounds generally known as IUPAC nomenclature.

The whole idea of the IUPAC nomenclature is to have a universally agreed pattern of naming organic compounds according to their structure.

The compound is always named in such a way that the substituents and the functional group receives the lowest number.

In this case, the functional group is alkyne hence the name of the compound ends in -yne. The methyl group is substituted on position four.

Putting all these together the compound should be named as;  4-methylhept-2-yne

For IUPAC nomenclature of an inorganic compound

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Answer:

Boiling

Explanation:

When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.

When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.

Answer: 8 for both

Explanation:

Explanation:

b. Bromo cyclopentane + HBr

Answer:

Choice A. [tex]\rm H_{2}[/tex] would be produced at the negative electrode.

Explanation:

Ionic equation for this reaction:

[tex]2\, {\rm K^{+}} + 2\, {\rm Cl^{-}} + {2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm K^{+}} + {\rm 2\, OH^{-}}[/tex].

Net ionic equation:

[tex]2\, {\rm Cl^{-}} + 2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm OH^{-}}[/tex].

Half-equations:

[tex]2\, {\rm Cl^{-}} \to {\rm Cl_{2}} + 2\, {e^{-}}[/tex].

(Electrons travel from the solution to an electrode.)

[tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex].

(An electrode supply electrons to the solution to reduce some of the [tex]\rm H[/tex] atoms from [tex]\rm H_{2}O[/tex].)

In a DC circuit, electrons always enter the circuit from the negative terminal of the power supply and return to the power supply at the positive terminal.

The negative electrode is connected to the negative terminal of the power supply. Electrons from the power supply would flow into the solution through this electrode.  

This continuous supply of electrons at the negative electrode would drive a reduction half-reaction. In this question, that corresponds to the reduction of water: [tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {\rm e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex]. Hence, [tex]\rm H_{2}[/tex] would be produced at the negative electrode.

Answer:

155 °C

Explanation:

Step 1: Given data

Step 2: Convert the initial temperature to Kelvin

We will use the following expression.

K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K

Step 3: Calculate the final temperature

Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 4.20 L × 296.2 K/2.91 L

T₂ = 428 K

Step 4: Convert the final temperature to Celsius

We will use the following expression.

°C = K - 273.15 = 428 - 273.15 = 155 °C

Answer:

Neutral

Explanation:

pKa of acid = -log Ka

                   = -log (1.8 x 10^-5)

                  = 4.74

pKb of base = -log Kb

                      = 4.74

pKa of acid = pKb of base

salt pH formula   : pH = 7 + 1/2 [pKa -pKb ]

here pKa = pKb

so pH = 7

the salt it is CH3COONH4 exactly neutral solution .

If a salt is formed by combining NH₃ (Kb=1.8×10⁻⁵) and CH₃COOH (Ka=1.8×10⁻⁵), an aqueous solution of this salt would be neutral.

pH of any solution tells about the acidity or basicity or neutral nature of the solution.

pH of any solution is directly proportional to the acid dissociation constant value (Ka) and base dissociation constant (Kb). In the question it is given that,

Value of Kb for NH₃ = 1.8×10⁻⁵

Value of Ka for CH₃COOH =  1.8×10⁻⁵

Ka & Kb values for the base and acid is same means it dissociates with same extent. So the aqueous solution of this acid and base is a neutral in nature as they have same number of acid and base ions in it.

Hence resultant solution will be a neutral solution .

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Answer:

An ecosystem encompasses living organisms and the nonliving elements of their environments. Hence, the components of an ecosystem include animals, plants, microorganisms, rocks, soil, minerals, atmosphere, and the surrounding water masses. An ecosystem can be huge, cutting across several nations, or it can be relatively small, such as the body of an animal, which is home to numerous microorganisms.

We can describe ecosystems under two headings – natural ecosystems and unnatural ecosystems. Unnatural ecosystems, which include agricultural and urban areas, are greatly modified and maintained by human activity. Conversely, natural ecosystems are self-sufficient, balanced ecological units, with a high proportion of native biodiversity and minimal human disruption.

The natural ecosystem is broad. It is divided into two major groups – terrestrial and aquatic ecosystems. These are further divided into many other smaller types of ecosystems as outlined in this article.

Explanation:

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Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

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Answer:

The answer is

Explanation:

1 mole of acid.

Hope this helps....

Have a nice day!!!!

A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.

A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.

In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.

A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.

Thus, 1 M in acid and base solution has the greatest buffer capacity.

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