Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH:
(a) gastric juice, pH= 1.80
(b) urine, pH 4.75 56.

Explanation:

The equation is given as;

2 A + B + 3 C → Products

The order of the reaction refers to the extent at which the rate depends n the concentration of the reactant.

The order of reaction is experimentally obtained. It can also be obtained from the rate law of the reaction.

If the rate law is given as;

rate law = k [A]²[B][C]³

Then the order is second order with respect to A.

The order is second order with respect to A.

Given that;

2A + B + 3C → Products at 470 K

Find:

Order of reaction with respect to A

Computation:

The reaction that takes place refers to how much the rate is influenced by the reactant concentration.

The reaction order is determined empirically. This can also be derived from the reaction's rate law.

Rate law = k[A]²[B][C]³

So, The order is second order with respect to A.

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Answer:

8.05 moles

Explanation:

5.50 / 2.954 = x / 4.325

x = 8.05

According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added  so that the balloon expands to 4.325 L.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.

In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314  so n=V/R=4.325/8.314=0.520 moles.

Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.

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Answer:

glue that would make the strongest craft stick tower.

Explanation:

Independent variable: In statistics and research methods, the term "independent variable" is determined as a variable that is being changed, controlled, or altered in an experiment or research by the researcher or the experimenter to see its effect on DV or dependent variable. However, it is said that independent variable directly effect the dependent variable.

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:

804 J

Explanation:

Step 1: Given data

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]

Q = 804 J

Answer:

OPTION C is correct

3.86 x 10-19 J

Explanation:

Energy of the line can be calculated using below formula

E= h ν.................(1)

Where E= energy

h= plank constant= 6.626 10-34 J s

c=speed of light=3 x 108 m/s

But we know that Velocity V= = c / λ

Then substitute into equation (1) we have

E = h c / λ.............(2)

We can calculate our( hc ) in nm for unit consistency

h c =( 6.626 ×10^-34)x(3×108)

h c = (1.986 x 10-16 )

hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)

E = h c / λ = (1.986 x 10-16) / 515

E = 3.86 x 10-19 J

Therefore, the Energy is 3.86 x 10-19 J

Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.  

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,  

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,  

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,  

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,  

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,  

74.91/15.999 = 4.68.  

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

Answer:

See figure 1

Explanation:

For this reaction, we have the production of a carbanion as the first step. The base "ethoxide" can remove a hydrogen-producing a negative charge in the carbon (enolate anion 1). Then this negative charge can attack the carbon of the carbonyl group in the molecule acetaldehyde and the tetrahedral intermediate 2 is form. In the next step, we have the protonation of the oxygen to produce alcohol 3. A continuation we have the hydrolysis of the ester groups to produce the Dicarboxyamino alcohol and finally, we have a decarboxylation reaction we will produce the amino acid Threonine.

To further explanations see figure 1

I hope it helps!

Answer:

(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants

Determining whether the statements about equilibrium is True or False

A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE

B) When a system is at equilibrium, Keq = 1 : TRUE

C) The rates of the forward reaction and the reverse reaction are equal at equilibrium :  TRUE

D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE

In a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.

A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).

Hence we can conclude that the answers to your questions are as listed above.

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Answer:

461.85 degrees Celsius

Answer:

[tex]Kp=5.14[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

Thus, the equilibrium expression is written as:

[tex]Kp=\frac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3}[/tex]

And in terms of the reaction extent:

[tex]Kp=\frac{(2x)^2}{(10-x)(10-3x)^3}[/tex]

Thus, from the equilibrium pressure of ammonia we can compute the reaction extent:

[tex]p_{NH_3}=2x=6.0 atm\\\\x=3.0atm[/tex]

Therefore, the equilibrium constant turns out:

[tex]Kp=\frac{(2*3.0)^2}{(10.0-3.0)(10.0-3*3.0)^3}\\\\Kp=5.14[/tex]

Regards.

Answer:

1. 0.00352 M

2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)

3. 0.00534 M

Explanation:

1.

Mass of strontium hydroxide= 10.45 g

Volume of solution = 41.00 ml

Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles

Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M

2.

2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)

3.

Concentration of acid CA= the unknown

Volume of acid VA= 31.5 ml

Concentration of base CB= 0.00352 M

Volume of base VB= 23.9 ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB = NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

CA= 0.00352 × 23.9 ×2/31.5 ×1

CA= 0.00534 M

A. The molarity of the Sr(OH)₂ solution is 2.09 M

B. The balanced equation for the reaction is

2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂O

C. The molarity of the acid, HNO₃ is 3.17 M

A. Determination of the molarity of the Sr(OH)₂ solution

Mass of Sr(OH)₂ = 10.45 g

Molar mass of Sr(OH)₂ = 88 + 2(16 + 1) = 122 g/mol

Mole = mass / molar mass

Mole of Sr(OH)₂ = 10.45 / 122

Mole of Sr(OH)₂ = 0.0857 mole

Volume = 41 mL = 41 / 1000 = 0.041 L

Molarity = mole / Volume

Molarity of Sr(OH)₂ = 0.0857 / 0.041

B. The balanced equation for the reaction.

C. Determination of the molarity of the acid, HNO₃.

From the balanced equation above,

The mole ratio of the acid, HNO₃ (nA) = 2

The mole ratio of the base, Sr(OH)₂ (nB) = 1

From the question given above,

Volume of base, Sr(OH)₂ (Vb) = 23.9 mL

Molarity of base, Sr(OH)₂ (Mb) = 2.09 M

Volume of acid, HNO₃ (Va) = 31.5 mL

MaVa / MbVb = nA/nB

(Ma × 31.5) / (2.09 × 23.9) = 2

(Ma × 31.5) / 49.951 = 2

Cross multiply

Ma × 31.5 = 49.951 × 2

Ma × 31.5 = 99.902

Divide both side by 31.5

Ma = 99.902 / 31.5

Thus, molarity of the acid, HNO₃ is 3.17 M

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Answer:

C;  ΔSuniv>0

Explanation:

In this question, we want to select which of the options must be true.

What we should understand is that for a process to be spontaneous, the change in entropy must be greater than 0 i.e the change in entropy must be positive.

Looking at the options we have; option C is the correct answer.

Option B looks correct but it is wrong. This is because if change in universal entropy is greater than zero, then change in Gibbs free energy must be less than zero for spontaneity to occur

It can be deduced that for a spontaneous process, B. ΔG>0.

It should be noted that a spontaneous process simply means a process that occurs without input of matter or electrical energy.

In this case, for a spontaneous process, it's true that ΔG>0, it should be noted that a spontaneous process related to the second law in thermodynamics. This is characterized by an increase in entropy.

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Answer:

See figure 1

Explanation:

We have to remember that in the isomer structures we have to change the structure but we have to maintain the same formula, in this case [tex]C_4H_1_1N[/tex].

In the formula, we have 1 nitrogen atom. Therefore we will have as a main functional group the amine group.

In the amines, we have different types of amines. Depending on the number of carbons bonded to the "N" atom. In the primary amines, we have only 1 C-H. In the secondary amines, we have two C-N bonds and in the tertiary amines, we have three C-N bonds.

With this in mind, we can have:

-) Primary amines:

1) n-butyl amine

2) sec-butyl amine including 2 optical isomers

3) isobutyl amine

4) tert-butyl amine

-) Secondary amines:

5) N-methyl n-propyl amine

6) N-methyl isopropyl amine

7) N, N-diethyl amine

-) Tertiary amines:

8) N-ethyl N, N-dimethyl amine

See figure 1

I hope it helps!

Answer:

Heat of vaporization refers to condensation and heat of fusion refers to melting.

Explanation:

Heat of vaporization or heat of evaporation, is defined as the energy required to convert liquid substance into a gas which creates condensation. As a gas condenses to a liquid, heat is released.

Heat of fusion refers to melting because heat of fusion is defined as the energy required to change any amount of substance when it melts.

Hence, the correct answer is "Heat of vaporization refers to condensation and heat of fusion refers to melting.".

Answer:

Yes

Explanation:

The balanced equation of the reaction is;

FeBr2 (aq) + Cl2 (g) → FeCl2 (aq) + Br2 (aq)

This reaction is possible because chlorine is more electronegative than bromine and can displace it from its salt.

In group seventeen, electro negativity decreases down the group. Hence as we move down the group, elements become less electronegative and can be displaced from their salt by more electronegative elements found earlier in the group.

Hence chlorine can displace bromine in FeBr2 to form FeCl2.

Answer:

Yes, because Cl2 has higher activity than Br2

Explanation:

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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Answer:

[tex]m_{CaCO_3}=0.179gCaCO_3[/tex]

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2[/tex]

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

[tex]m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3[/tex]

Regards.

The mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g

From the question,

We are to determine the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP.

First, we will determine the number of mole of CO₂ required to be produced

From the formula

PV = nRT

Where

P is the pressure

V is the volume

n is the number of moles

R is the ideal gas constant

and T is the temperature

Then, we can write that

[tex]n = \frac{PV}{RT}[/tex]

From the question,

V = 40.0 mL = 0.04 L

At STP

P = 1 atm

T = 273.15 K

and

R = 0.08206 L atm mol⁻¹ K⁻¹

Putting the parameters into the formula, we get

[tex]n = \frac{1 \times 0.04}{0.08206 \times 273.15}[/tex]

∴ n = 0.0017845 mole

Now, we will write the balanced chemical equation for the decomposition of CaCO₃

CaCO₃ → CaO + CO₂

This means,

1 mole of CaCO₃ will decompose to produce 1 mole of CO₂

Since 0.0017845 mole of CO₂ is to be produced,

Then,

0.0017845 mole of CaCO₃ would be required

Now, for the mass of CaCO₃ required,

Using the formula

Mass = Number of moles × Molar mass

Molar mass of CaCO₃ = 100.0869 g/mol

∴ Mass of CaCO₃ required = 0.0017845 × 100.0869

Mass of CaCO₃ required = 0.178605 g

Mass of CaCO₃ required ≅ 0.179 g

Hence, the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g

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Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.

Molality can be defined as the mass of solute present in 1000 grams of solvent.

Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]

The moles of NaCl has been calculated as;

3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]

3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]

Moles of NaCl = 3.140 [tex]\times[/tex] 2.692

Moles of NaCl = 8.45288 mol

The moles can be expressed as;

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Molecular weight of NaCl = 58.5 g/mol

The mass of NaCl can be calculated as:

8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]

Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol

Mass of NaCl = 494.493 grams.

The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.

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Answer:

.217, .311, and .472, respectively.

Explanation:

The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).

X of helium=3.50/16.10 = .217

X of krypton=5.00/16.10 = .311

X of neon=7.60/16.10 = .472

Answer:

See explanation.

Explanation:

Hello,

In this case, we can realize that water has a very simple atomic structure which consists of two hydrogen atoms bonded to one oxygen atom. The nature of the atomic structure of water causes its molecules to have unique electrochemical properties. The hydrogen side of the water molecule has a slight positive charge whereas at the other side of the molecule a negative charge exists. This molecular polarity causes water to be a powerful solvent and is responsible for its strong surface tension.

Moreover, water is involved in several both inorganic and organic chemical reactions leading to hydration, for example, the conversion of alkenes to alcohols, the hydrolysis of acyl halides, anhydrides, esters and amides to carboxylic acids and the hydration of a raft of inorganic salts that exist as hydrates only, such as copper (II) sulfate pentahydrate and so.

Best regards.

Answer:158 days (D)

Explanation:

It will take 158 days for grass contaminated with Sr-90 to be safe for cattle to eat. Therefore, option (A) is correct.

The half-life of a radioactive material is defined as the time that is needed to reduce the initial quantity of a radioactive element to half after disintegration.

The half-life of a radioactive element can be described as the characteristic of the element and does not influence by the initial amount of the radioactive substance.

Given, the half-life of the Strontium-90 = 28 days

The rate constant of the decay can be determined as:

[tex]t_{\frac{1}{2} } =\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{t_{\frac{1}{2} } }[/tex]

k = 0.693/28

k = 0.025 day⁻¹

The concentration of Strontium-90 reduced to less than 2% is safe. Therefore final concentration [A] = 2 % = 0.02

[tex]t = \frac{2.303}{k} log \frac{[A_o]}{[A]}[/tex]

[tex]t = \frac{2.303}{0.02475} log \frac{1}{[0.02]}[/tex]

t = 158 days

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Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

Answer:

[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:

[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]

Now we can balance the reaction:

[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):

[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]

[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]

In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].

With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]

We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]

I hope it helps!

Answer:

73.1707317073%

=> Approximately 73.2%

Explanation:

Total = 60 + 10 + 12

=> 82

Test tubes are 60/82

=> 30/41

=> 73.1707317073%

=> Approximately 73.2%

Answer:

73.17%

Explanation:

To find the percentage of test tubes to the overall glassware, we need to get the number of test tubes divided by the total number of glassware.

12 beakers + 10 flasks + 60 test tubes = 82 glassware

% test tube = 60 / 82 = .7317 ==> 73.17 %

So 73.17 % of the glassware was test tubes.

Cheers.

Answer:

Answers are in the explanation.

Explanation:

In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:

Entropy of gases >>> entropy of liquid > entropy of solids.

The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.

In the reactions:

1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)

As 1 gas is produced, entropy of products is higher than entropy of reactants. That means  ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)

2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.

3. Mg(s) + Cl₂(g) → MgCǐ₂(s)

You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

4. SO₃(g) + H₂O(I) → H₂SO₄(I)

1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:

1. ΔSsys > 0.

2. ΔSsys < 0

3. ΔSsys < 0

4. ΔSsys < 0

Recall:

Thus:

In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.

In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.

In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.

In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.

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Answer:

Explanation:

Answer in attached file .

Answer:

e is the suitable answer for that. I think it is correct.

Answer: The correct option is E ( the reactant that is completely used up by a reaction).

Explanation:

A LIMITING REACTANT can be defined as the reagent or the substance that is involved in a chemical reaction which determines when the reaction will stop. This is because it is COMPLETELY used up in the reaction. Reactants that are called limiting reactants is because the quantity of these reagents are capable of limiting the amount of products formed. And by doing so, the chemical reaction cannot proceed further with the absence of this reactant. Using an attached diagram below to illustrate further:

The reagents D and E reacts to form F as the product. In this reaction, reactant E is the limiting reagent because there is still some left over D in the products. Therefore, D was in excess when E was all USED UP.

Therefore the CORRECT option is E which states that the reactant that is completely used up by a reaction, best describes a limiting reactant.

Option A is WRONG because it's the concentration of both reactants in chemical equation can limit the speed of that reaction.

Option B is WRONG because it's when the concentration of a particular reactant is either increased or decreased can affect the position of equilibrium.

Option C and D are wrong because the reactant that remains in the end of a reaction and can produce the greatest amount of product is the one in EXCESS.

Answer:

Underneath the frozen upper layer, the water remains in its liquid form and does not freeze. Also, oxygen is trapped beneath the layer of ice. As a result, fish and other aquatic animals find it possible to live comfortably in the frozen lakes and ponds. ... This irregular expansion of water is called anomalous expansion.

Explanation:

so pretty much its because there is water under the top frozen layer and air is trapped underneath the ice

Answer:

The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]

Explanation:

Given that:

mass of an unknown sample [tex]m_3[/tex] = 0.0850

temperature of the unknown sample [tex]t_{unknown}[/tex] = 100° C

initial temperature of the calorimeter can = 19° C

mass of copper [tex]m_1[/tex] = 0.150 kg

mass of water [tex]m_2[/tex]= 0.20 kg

the final temperature of the calorimeter can = 26.1° C

The objective is to compute the specific heat of the sample.

By applying the principle of conservation of energy

[tex]Q = mc \Delta T[/tex]

where;

[tex]Q_1 +Q_2 +Q_3 = 0[/tex]        

i.e

[tex]m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2+m_3 c_3 \Delta T_3 =0[/tex]

the specific heat capacities of water and copper are 4.18 × 10³ J/kg.K and 0.39 × 10³ J/kg.K respectively

the specific heat of the sample [tex]c_3[/tex] can be computed by making [tex]c_3[/tex]  the subject of the above formula:

i.e

[tex]c_3 = \dfrac{m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2}{m_3 c_3 \Delta T_3}[/tex]

[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (26.1 -19) + 0.20 \times 4.18 \times 10^3 \times (26.1 -19) }{0.0850 \times (100-26.1 )}[/tex]

[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (7.1) + 0.20 \times 4.18 \times 10^3 \times (7.1) }{0.0850 \times (73.9)}[/tex]

[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]

[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]

[tex]c_3 = \dfrac{6350.95}{6.2815}[/tex]

[tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]

The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]