The temperature of a 12.58 g sample of calcium carbonate [CaCO3(s)] increases from 23.6 Celsius to 38.3 Celsius. If the specific heat of CaCO3 is 0.82 J/g-K, how many joules of heat are absorbed during this process

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

Answer:

The answer is " 10.39"

Explanation:

Calculating acid moles:

[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]

Calculating NaOH moles:

[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]

calculating excess in OH-  Moles:

[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]

calculating total volume:

[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]

[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]

           [tex]=0.00025 M[/tex]

[tex]pOH = - \log 0.00025[/tex]

        = 3.6

[tex]pH = 14 - pOH[/tex]

      = 10.39

Answer:

3.07 × 10⁻⁴

Explanation:

Step 1: Calculate the concentration of H⁺

We will use the definition of pH.

[tex]pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M[/tex]

Step 2: Calculate the concentration of HY

5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:

[tex]Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M[/tex]

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}[/tex]

Answer:

Intermediate.

Explanation:

Hello,

In this case, we can rewrite the steps as:

[tex]A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)[/tex]

Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).

Best regards.

Answer:

D. [HCHO₂] > [NaCHO₂]

Explanation:

Formic acid, HCHO₂, is a weak acid that, in presence of its conjugate base, NaCHO₂ (CHO₂⁻), produce a buffer following H-H equation:

pH = pKa + log [CHO₂⁻] / [HCHO₂]

As pKa of the acid is 3.74 and pH of the solution is 3.11:

3.11 = 3.74 + log [CHO₂⁻] / [HCHO₂]

-0.63 = log [CHO₂⁻] / [HCHO₂]

0.2344 = [CHO₂⁻] / [HCHO₂]

A ratio [CHO₂⁻] / [HCHO₂] < 1, means:

Answer:

Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.

Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.

The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.

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The given question is incomplete, the complete question is:

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol

A) –309 kJ/mol

B) –329 kJ/mol

C) None of the above

D) –349 kJ/mol

E) –369 kJ/mol

Answer:

The correct answer is option D, that is, -349 kJ/mol.

Explanation:

Based on the given information, the reaction is:  

NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.  

Therefore, there is a need to reverse the reaction and change the sign on ΔG.  

Now the reaction will become,  

Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).  

Answer: A

Explanation:

Answer:

How slow or fast a reaction progresses

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

Answer:

Samarium:

Electron configuration:

Samarium

Explanation:

Samarium is a chemical element that belongs to the lanthanoid series. The lanthanoids are the chemical elements that follow lanthanum. They are all known to possess 4f orbitals. The 4f electrons are found in the antepenultimate shell of the elements of the lanthanoid series and they do not take part in chemical bonding. They are neither removed in bonding nor do they take part in crystal field stabilization of lanthanoid complexes.

The electronic configuration of samarium is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6 while the condensed, short hand electronic configuration is [Xe] 4f6 6s2. This corresponds to (noble gas]ns2(n − 2) f6 as required by the question, hence the answer provided above.

Answer:

pH = 5.24

Explanation:

Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.

Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:

[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M

[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.76M] / [0.24M]

Answer:

pH = 4.05

Explanation:

The pH of the benzoic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.187

pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]

Where [] can be understood as moles of each specie.

Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.

Initial moles:

Initial moles of benzoic acid and sodium benzoate are:

Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid

Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate

Moles after reaction:

Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:

HCl + C₆H₅COONa → C₆H₅COOH + NaCl

That means after reaction moles of both species are:

Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles

Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles

Replacing in H-H equation:

pH = 4.187 + log [0.0475] / [0.065]

pH = 4.05

Answer:

CoSO₄.7H₂O

Explanation:

Some salts are in its hydrate form to stabilize them. The hydrated form of CoSO₄ is:

CoSO₄.XH₂O

To find the hydration of the COSO₄ you must find the ratio of H₂O and CoSO₄ that is:

0.068921 moles H₂O / 0.0098459 moles CoSO₄ = 7

That means you have 7 moles of water per mol of CoSO₄ and the formula is:

Answer:

Solid Wood

Explanation:

Wood is like a solid block, whereas gases flow freely and liquids spread to fill the shape of their container.

Please let me know if I misunderstood the question, by the way.

Answer:

Concentration (C) = number of moles (n) / volume (v)

Therefore

number of moles (n) = concentration × volume

Concentration = 6.50M

Volume =0.175 L = 0.175dm³

n = 6.50 × 0.175

n = 1.138 moles

Moles of ammonia in the solution is

1.138 moles.

Hope this helps

Answer:

Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.  

Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.  

Answer:

Fronts

Explanation:

For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!

Hope this helps! Please mark as brainiest!

Answer:

[tex]Q=4154J[/tex]

Explanation:

Hello,

In this case, the involved heat in this heating process is considered to be computed via:

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

Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:

[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]

Regards.

Answer:

[tex]\large \boxed{\text{0.195 mol/L}}[/tex]

Explanation:

(a) Balanced equation

2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O  

(b) Moles of Ca(OH)₂

[tex]\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}[/tex]

(c) Moles of HNO₃

The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂

[tex]\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}[/tex]

(d) Molar concentration of HNO₃

[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the Ca(OH)$_{2}$ is $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]

Answer:

The freezing point of the solution is -1.4°C

Explanation:

Freezing point decreases by the addition of a solute to the original solvent, freezing point depression formula is:

ΔT = kf×m×i

Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.

Molality of the solution is:

-moles CaBr₂ (Molar mass:

189.9g ₓ (1mol / 199.89g) = 0.95 moles

Molality is:

0.95 moles CaBr₂ / 3.75kg water = 0.253m

Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:

CaBr₂(s) → Ca²⁺ + 2Br⁻

3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.

Replacing:

ΔT = kf×m×i

ΔT = 1.86°C/m×0.253m×3

ΔT = 1.4°C

The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,

Answer:

THE FREEZING POINT IS -1.41 °C

Explanation:

Using the formula of change in freezing point:

ΔTf = i Kf m

i = 3 (1 Ca, 2 Br)

i is the number of the individual elements in the molecules

Kf of water = 1.86 °C/m

mass of CaBr2 = 189.9 g

Calculate the Molar mass of CaBr2:

Molar mass  = ( 40 + 80*2) = 200 g/mol

Calculatee the molarity:

molarity = 189.9 g * 1 mole / 200 g/mol / 3.75 kg of water

molarity = 0.2532 M

So therefore, the change in freezing point is:

ΔTf = 1 Kf * M

ΔTf = 3 * 1.86 * 0.2532

ΔTf = 1.41 °C

The freezing point = old freezing point - change in freezing point

The freezing point = 0 - 1.41 °C = - 1.41 °C

The freezing point therefore is -1.41 °C

Answer:

The correct answer would be - 2.4KJ or, 2400J

Explanation:

Given:

heat capacity of liquid water - 4.18 J/g·°C

heat of vaporization - 40.7 kJ/mol

Mass of water = 1g

Moles of water = mass/molar mass

= 1g/18.016g

= 0.055 moles

Then,

Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)

= m *s*dt + moles * heat of vaporization

= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ

= 137.94J + 2.26KJ

=0.138KJ + 2.26KJ

=2.4KJ or, 2400J

Thus, the correct answer would be - 2.4KJ or, 2400J

Answer:

Explanation:

The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

For the steps below, refer to the ICE chart above.

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

Answer:

2.1 atm

Explanation:

Before we get the total pressure, we have to ensure all the gases have the same pressure unit.

Nitrogen gas = 0.75 atm

Helium = 490mmHg

To convert mmHg to atm;

760 mmHg = 1 atm

490 = x

x = 460 / 760 = 0.645 atm

Neon = 520 torr

Converting torr to atm;

760 torr = 1 atm

520 torr = x

x = 520 / 760 = 0.684 atm

The total pressure is then given as;

0.75 + 0.684 + 0.645 = 2.1 atm

Answer:

The solution is basic and after the equivalence point

Explanation:

HF reacts with NaOH as follows:

HF + NaOH → H₂O + NaF

1 mole of acid reacts per mole of base.

The solution of HF contains:

40.0mL = 0.0400L × (0.65mol / L) = 0.026 moles of HF,

The titration will reach equivalence point when you add 0.026 moles of NaOH. If more NaOH is added the solution will be basic because NaOH (A base) is in excess doing the solution basic.

Moles added of NaOH are:

0.400L × (0.100 mol / L) = 0.0400 moles of NaOH

As the addition of NaOH is > 0.026 moles,

Answer:

Option C. Electrons are shared between two atoms

Explanation:

Covalent bonding is a type of bonding which exist between two non metals.

In this bonding, electrons are shared between the two atoms involved in order to attain a stable octet configuration.

This can be seen when hydrogen atom combine with chlorine atom to form hydrogen chloride as shown below:

H + Cl —> HCl

Hydrogen has 1 electron in it's outmost shell and it requires 1 electron to attain a stable configuration.

Chlorine has 7 electrons in it's outmost shell and requires 1 electron to attain a stable configuration.

During bonding, both hydrogen and chlorine will contribute 1 electron each to form bond, thereby attaining a stable configuration. The bond formed in this case is called covalent bond as both atoms involved shared electron to attain a stable configuration.

C. Electrons are shared between two atoms.

What is Covalent Bonding?

For example:

In H₂ molecule; there is a covalent bond formation between two hydrogen atoms as the electron from each hydrogen atom is shared leading to the formation of hydrogen molecule.

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Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.

The balanced chemical reaction is:

[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]

Answer:

i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances

Explanation:

Answer:

None

Explanation:

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