Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =

Answer:

ITS NOT D. ITS B. 4.52x10^-9 M

Explanation:

Answer:

4.52 ×10–9 M

Explanation:

answer is oxygen .

oxygen is an exception in octet rule

Among the following given elements,oxygen is an element which cannot have an expanded octet.

Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds  are not less common  and are of equal stability as the compounds which obey octet rule.

Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.

Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does  not have the 2d -orbitals   due to which it  does not fulfill the criteria of an element to have an expanded octet.

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We have a solution of NaOH and H₂CO₃

First, NaOH will dissociate into Na⁺ and OH⁻ ions

The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃

The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule

If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃

The concentration of a solution is the amount of solute in a solution.

We have from the question that he mass of manganese = 1.584 g

Hence;

Amount of [tex]Mn^2+[/tex] = 1.584 g/55g/mol = 0.0288 moles

Recall that;

Number of moles = concentration * volume

Let the concentration of the solution be C

0.0288 moles = C * 1 L

C = 0.0288 moles/ 1 L

C= 0.0288 mol/L

Hence concentration of stock solution = 0.0288 mol/L

For solution A

From the dilution formula;

C1V1 = C2V2

where;

C1 = initial concentration

C2 = final concentration

V1 =  initial volume

V2= final volume

C1 = 0.0288  mol/L

V1 = 50.00 mL

C2 = ?

V2 = 1000.0 mL

C2 = 0.0288  mol/L * 50.00 mL/1000.0 mL

C2 = 0.00144 mol/L

Hence, concentration of solution A is 0.00144 mol/L

For solution B

C1V1 = C2V2

C1 = 0.00144 mol/L

V1 = 10.00 mL

C2 = ?

V2 = 250.0 mL

C2 = 0.00144 mol/L * 10.00 mL/250.0 mL

C2 = 0.0000576 mol/L

Hence, concentration of solution B is 0.0000576 mol/L

For solution C

C1 = 0.0000576 mol/L

V1 = 10.00 mL

C2 = ?

V2 = 500.0 mL

C2 = 0.0000576 mol/L * 10.00 mL/500.0 mL

C2= 0.000001152 mol/L

Hence, concentration of solution C is 0.000001152 mol/L

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Answer:

96 mmHg

[tex]h=96mmHg[/tex]

Explanation:

From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .

The difference of height of the mercury level is given as,

h=9.60cm

h=9.60(10mm/1cm)

[tex]h=96mm[/tex]

But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height

P(gas)=h

P(gas)=96mmHg

This pressure is as a result of the presence of gas.

Therefore, the pressure of the argon gas in the container is 96mmHg.

The pressure of the argon in the container was 96mmHg.

We were told that the manometer has closed ends which means that the

height will not be affected by atmospheric pressure.

The height which is the difference in mercury level is

h=9.60cm

We can convert it to millimeter by multiplying it by 10

h=9.60 × 10 = 96mm

The pressure of the closed end manometer will be equal to the height

P(gas)=h

P(gas)=96mmHg

The pressure of the argon gas in the container is 96mmHg.

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Explanation:

Carbon-12 atoms have stable nuclei because of the 1:1 ratio of protons and neutrons.

Carbon-14 atoms have nuclei which are unstable. C-14 atoms will undergo alpha decay and produce atoms of N-14. Carbon-14 dating can be used to determine the age of artifacts which are not more than 50,000 years old.

Answer:

C.

Explanation:

I believe that it should be A and B.

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]

Whose equilibrium expression is:

[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

Explanation:

Entropy refers to the degree of disorderliness of a system.

a. A snowman melts on a spring day.

Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.

b. A document goes through a paper shredder.

Entropy increases because random, disorganized bits of paper are left.

c. A water bottle cools down in a refrigerator.

Entropy  decreases because temperature is directly proportional to entropy.

d. Silver tarnishes

Entropy increases because random bits of the sliver particles are formed.

e. Dissolved sugar precipitates out of water to form rock candy.

Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.

Answer:

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]

In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:

[tex]T_2=T_1+\frac{Q}{mCp}[/tex]

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]

Thereby, answer is B. 58°C .

Regards.

The kinetic-molecular theory of gases assumes that ideal gas molecules

(1) are constantly moving;

(2) have negligible volume;

(3) have negligible intermolecular forces;

(4) undergo perfectly elastic collisions; and

(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.

The five main postulates of the KMT are as follows:

(1) the particles in a gas are in constant, random motion,

(2) the combined volume of the particles is negligible

(3) the particles exert no forces on one another,

(4) any collisions between the particles are completely elastic.

(5) the average kinetic energy of the particles is proportional to the temperature in kelvins.

Answer:

gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata

Explanation:

vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs

Answer:

Covalent bonding and non-covalent bonding

Answer:

well, first off. the formula for carbon tetrachloride is CCl4

We need to find the molar mass of carbon tetrachloride

1(Mass of C) + 4(mass of chlorine)

1(12) + 4(35.5)

12 + 142

154 g/mol

Number of moles of CCl3 in 543.2g CCl3

n = given mass / molar mass

n = 543.2/153

n = 3.53 moles

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Answer:

3.53 moles

Explanation:

Answer:

C. Gas

Explanation:

Explanation:

Antacids are medications used to manage the symptoms of indigestion and heartburn. Antacids contain active ingredients that are bases. These allow antacids to neutralize any stomach acid which could be causing digestive discomfort.

Answer:

b

Explanation:

Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.

The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.

The correct option is b.

Answer:

concentration   in    mmol/L = 8.97 × 10⁻¹ mmol/L

Explanation:

Given that:

the number of moles of  silver(I) nitrate(AgNO3) the chemist used in preparing a solution = 269 mmol = 269 × 10⁻³ mmol

The volume of the volumetric flask = 300 mL  = 300 × 10⁻³ L

In order to calculate the concentration in mmol/L of the chemist's silver(I) nitrate (AgNO3) solution , we used the formula which can be expressed as;

[tex]concentration \ in \ mmol/L = \dfrac{ number \ of \ mmol}{vol. \ of \ the \ solution}[/tex]

[tex]concentration \ in \ mmol/L = \dfrac{ 269 * 10^{-3 } \ mmol }{300 * 10^{-3} \ L }[/tex]

concentration   in    mmol/L = 0.8966   mmol/L

concentration   in    mmol/L = 8.97 × 10⁻¹ mmol/L

Answer:

Answers are in the explanation

Explanation:

It is possible to obtain K of equilibrium of related reactions knowing the laws:

A + B ⇄ C K₁

C ⇄ A + B K = 1 /K₁

The inverse reaction has the inverse K equilibrium

2A + 2B ⇄ 2C K = K₁²

The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients

For the reaction:

2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K

1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)

This is the inverse reaction but also the coefficients are dividing in the half, that means:

[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]

2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)

Here,the only change is the coefficients are the half of the original reaction:

[tex]K_2 = K^{1/2}[/tex]

3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)

This is the inverse reaction. Thus, you have the inverse K of equilibrium:

[tex]K_3 = \frac{1}{K}[/tex]

Answer:

The compound contains a o molecular orbital formed by the overlap of one carbon sp2 hybrid orbital and one hydrogen sp3 hybrid orbital.

Explanation:

Molecular orbital is function which describes wave like behavior of an electron in a molecule. The molecular orbital theory describes the electronic structure of molecule using quantum mechanics. Electrons are not assigned to individual bonds between atoms. The compound contains sp2 hybrid orbial which is polar C - CI bond.

Answer:

Kindly check the explanation section.

Explanation:

PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.

The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and

Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.

In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.

The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.

In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.

The expected structure of the NaCl would be the image attached below.

The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.

For more information about the structure of NaCl, refer to the link:

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Answer:

12.5g

Explanation:

Half life = 2.4 Minutes.

The half life of a compound is the time it takes to decay to half of it's original concentration or mass.

Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)

Initial mass = 100g

First half life;

100g --> 50g

Second half life;

50g --> 25g

Third half life;

25g --> 12.5g

The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]

Thus, 3 half-lives has elapsed.

Original amount (N₀) = 100 g

Number of half-lives (n) = 3

[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]

Thus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g

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Answer:

3-pentanol

Explanation:

In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].

Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at  all). So, the structures only can be primary and secondary structures.

With this in mind, our options are:

-) 1-pentanol

-) 2-pentanol

-) 3-pentanol

Now we can analyze each structure:

-) 1-pentanol

The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).

-) 2-pentanol

On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).

-) 3-pentanol

In this structure, we have the "OH"  bonded to carbon three. So, we can analyze each fragment:

   -) m/z 59

This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.

   -) m/z 71

This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.

   -) m/z 73

This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.

See figure 2

I hope it helps!

Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

Total volume = 100 + 50 = 150 mL

Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

So therefore, the heat evolved during the reaction is:

Heat = 150 * 4.18 * ( 31.4 - 22.3)

Heat = 150 * 4.18 * 9.1

Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)  

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\?\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]?=\frac{3kg*11mg}{1kg}\\ \\?=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

there will be no chemical reaction

Answer:

See explanation

Explanation:

The dissolution of oxygen in water is exothermic. Hence, at higher temperature, less oxygen is dissolved in water.

If the chemist commits an error and records 27°C instead of 21°C, the dissolved oxygen concentration will be found to be lower than it should be at 27°C because dissolution of gases in water is exothermic.

Explanation:

they are called hydrocarbyls

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Answer:

The first carbon atom that attaches to the functional group is referred to as the alpha carbon.

answer should be 1.4 cm³

1 L = 10 and so

dL = 100 and then

cL = 1,000

mL = 0.001 m³

1 m³ = 1,000

dm³ = 1,000,000

cm³ = 1,000,000,000

mm³ = 1,000 L

So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL

1,400 mL = 1,400 cm³ = 1.4 L = 140 cL

Answer:

1.4 cm^3

Explanation:

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]