Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4

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:

30.12 mL.

Explanation:

We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:

Phosphoric acid H3PO4 will dissociate in water as follow:

H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)

From the balanced equation above,

1 mole of H3PO4 produces 3 moles of H+.

Therefore, XM H3PO4 will produce 8.70 M H+ i.e

XM H3PO4 = 8.70/3

XM H3PO4 = 2.9 M.

Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.

Next, we shall write the balanced equation for the reaction. This is illustrated below:

2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, H3PO4 (nA) = 2

Mole ratio of the base, Ba(OH)2 (nB) = 3

Data obtained from the question include the following:

Molarity of base, Ba(OH)2 (Mb) = 6.50 M

Volume of base, Ba(OH)2 (Vb) =.?

Molarity of acid, H3PO4 (Ma) = 2.9 M

Volume of acid, H3PO4 (Va) = 45 mL

The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.9 x 45 / 6.5 x Vb = 2/3

Cross multiply

2 x 6.5 x Vb = 2.9 x 45 x 3

Divide both side by 2 x 6.5

Vb = (2.9 x 45 x 3) /(2 x 6.5)

Vb = 30.12 mL

Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL

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 MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C

Explanation:

1604 J of heat is added to 48.9 g of hexane

To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.

Since 1604 J of heat = 48.9 g of hexane

Molar mass of hexane = 86 g/mol = 1 mole

then;

1604 J = 48.9 g

x = 86 g

x = 1604 * 86 / 48.9

x = 4205.4 J

Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.

In other words,

heat = molar heat capacity * temperature change

molar heat capacity = heat/ temperature change

Molar heat capacity = 4205.4 J / 14.5 C

Molar heat capacity = 290.027 J/C

The molar heat capacity of hexane is 290.027 J/ C

Answer:

Companies know that people will be willing to spend more to get an in-demand product.

Explanation:

When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.

Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.

Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.

Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.

Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.

Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.

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Answer:

6.5 mg/L.

Explanation:

Step one: write out and Balance the chemical reaction in the Question above:

NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.

Step two: Calculate or determine the number of moles of AgCl.

So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:

Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.

Step three: Calculate or determine the number of moles of NiCl2.

Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.

Step four: detemine the mass of NiCl2.

Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.

Step five: finally, determine the concentration of NiCl2.

1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.

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:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

[tex]left\ over=0g[/tex]

Regards.

Answer:

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Explanation:

The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.

For the reaction between iron and copper II nitrate, the molecular reaction equation is;

Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)

Ionically;

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Answer:

The concentration in mol/L =  4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L =  4.34188 mol/L

The concentration in mol/L =  4.342 mol/L

Answer:

-434.14 kJ

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Calculate the standard free energy change (ΔG°r) for the reaction

We will use the following expression.

ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))

ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)

ΔG°r = -959.42 kJ

Step 3: Calculate the standard free energy change for 1.81 moles of NH₃

959.42 kJ are released per 4 moles of NH₃.

[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]

Answer:

-69 kJ

Explanation:

Step 1: Given data

Standard cell potential (E°cell): +0.24 V

Electrons involved (n): 3 mol

Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell

We will use the following expression.

ΔG° = -n × F × E°cell

where,

F is Faraday's constant (96,485 C/mol e⁻)

ΔG° = -n × F × E°cell

ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V

ΔG° = -69 kJ

Answer : HBr polar

please add me to brainalist

Answer:

a. Sodium cyclopentanecarboxylate

b. No reaction

Explanation:

In this case, in the cyclopentanecarboxylic acid we have a carboxylic acid functional group. Therefore we have an "acid". The acids by definition have the ability to produce hydronium ions ([tex]H^+[/tex]).

With this in mind, for molecule a. we will have an acid-base reaction, because NaOH is a base. When we put together an acid and a base we will have as products a salt and water. In this case, the products are  Sodium cyclopentanecarboxylate (the salt) and water.

For the second molecule, we have the hydronium ion  ([tex]H^+[/tex]). This ion can not react with an acid. Because, the acid will produce the hydronium ion also, so a reaction between these compounds is not possible.

See figure 1

I hope it helps!

Answer:

2

Explanation:

A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.

Answer:

the answer is 2

Explanation:

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to:

Answer:

A) 3.17 g of Zn

Explanation:

Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.

Zn⁺²(aq) + 2e⁻ → Zn(s)

We can establish the following relations:

The mass of Zn deposited under these conditions is:

[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]

Answer:

A.) 3.17

Explanation:

I got it right in class!

Hope this Helps!! :))

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:

A divergent plate boundary  

Explanation:

At a divergent boundary, the plates pull away from each other and generate new crust.

Answer:

Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.

Explanation:

PLATO ANSWER

Answer:

2.81 g of H2O.

Explanation:

We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.

This can be obtained as follow:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.

1 mole of O2 = 16x2 = 32 g.

Thus 6.022×10²³ molecules is present in 32 g of O2,

Therefore, 1.25×10²³ molecules will be present in =

(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.

Therefore, 1.25×10²³ molecules present in 6.64 g of O2.

Next, the balanced equation for the reaction. This is given below:

CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)

Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.

This can be obtained as follow:

Molar mass of CH4 = 12 + (4x1) = 16 g/mol.

Mass of CH4 from the balanced equation = 1 x 16 = 16 g

Molar mass of O2 = 16x2 = 32 g/mol.

Mass of O2 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol.

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2.

Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.

From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.

Therefore, CH4 is the limiting reactant.

Finally, we shall determine the mass of H2O produced from the reaction.

In this case, the limiting reactant will be used because it will give the maximum yield of H2O.

The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted to produce produce 36 g if H2O.

Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.

Therefore, 2.81 g of H2O were obtained from the reaction.

The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.

Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.

Given chemical reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Moles of CH₄ will b calculate as:

n = W/M, where

W = given mass = 1.25g

M = molar mass = 16g/mol

n = 1.25/16 = 0.078 moles

Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³

Given molecules of O₂ = 1.25×10²³

Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.

From the stoichiometry of the reaction it is clear that:

1 mole of CH₄ = will produce 2 moles of H₂O

0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O

Mass of H₂O will be calculated by using its moles as:

W = (0.156)(18) = 2.8g

Hence required mass of H₂O is 2.8g.

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Answer:

- Addition of Ba(OH)2: favors the formation of a precipitate.

- Undergo a chemical reaction forming soluble species.

- Addition of CuSO4 : favors the formation of a precipitate.

Explanation:

Hello,

In this case, since the dissociation reaction of barium sulfate is:

[tex]BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)[/tex]

We must analyze the effect of the common ion:

- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

- By adding sodium nitrate, the following reaction will undergo:

[tex]BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)[/tex]

So the precipitate will turn into other soluble species.

- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

All of this is supported by the Le Chatelier's principle.

Best regards.

Answer:

C. Electron cloud

the electron is around 1/2000 times the size of the proton.

If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row

The two layers are part of the thermosphere are  exosphere and ionosphere.

The Exosphere is the topmost layer of the Earth's atmosphere.

and its gradually disappear into the vacuum of space.

It consist two parts that are:

exosphere and ionosphere.

Thus, option "A" is correct, the rest of the option is not a part of thermosphere.

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Answer:

hi hope your doing great the answer is A

Explanation:

its on Edge 2020

hope i helped :)

Answer:

1.8x10¹⁷ molecules of CO are in each breath we take

Explanation:

Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.

A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.

In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:

2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =

[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take

A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.

Now CO molecules in each breath is

[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]

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Answer:

16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.

Explanation:

A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

The solution is basic since it's pH lies in the basic region.

Hope this helps you

Answer:

The correct answer is - 1.02 V

Explanation:

From the reduction-oxidation reaction:

Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:

Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s)                    Eº= ‑0.25 V

Oxidation (anode) :  2 x (Fe²⁺ → Fe³⁺ + e-)(aq)                Eº= -0.77 V

                                -------------------------------------

                     Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):

Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V

The words given are not clear, so the clear question is as follows:

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:

A. β-1,4- and α-1,6-glycosidic

B. α-1,4-glycosidic

C. α-1,4-galactose

D. an unbranched glucose

E. a branched fructose

F. α-1,6-glycosidic

Amylose is ......... polymer of ....... units joined by ........ bonds.

Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answer:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Explanation:

Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.

Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.

Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.

Hence, the correct answers in the sequential order are:

Amylose:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

Amylopectin:

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Answer:

Explan ionization energy, atomic radius, and electron affinityation:

This question most likely has answer choices. The possible answer choices are as followed:

The answers are Ionic radii tend to increase down a group, Cationic radii tend to decrease across a period, and Ionic radii increase when switching from cations to anions in a period (1st, 4th, and 5th options).

Answer:

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Explanation:

Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:

Q = c * m * ΔT

Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).

In this case:

Replacing:

[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]

Q= 64,121.2 J

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J