Calculate number of silver atoms in
pure,
silver bracelet that
mass of 330, CAg=108. Na = 6.02x 10²3)
has
a​

Answers

Answer 1

Answer:

[tex]1.84x10^{24}atoms[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the number of silver atoms in 330 grams of a silver bracelet by knowing 1 mole of silver has a mass of 108 grams and also contains 6.022x10²³ atoms; thus, the mathematical setup is shown below:

[tex]330g*\frac{1mol}{108g}*\frac{6.022x10^{23}atoms}{1mol}[/tex]

Which is equal to:

[tex]1.84x10^{24}atoms[/tex]

Regards!


Related Questions

All bases dissociate
True or false

Answers

Answer:

verdadero

Explanation:

porque esoo [tex]\lim_{n \to \infty} a_n x_{123} \frac{x}{y} \sqrt[n]{x} x^{2} \sqrt{x} \pi \neq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right.[/tex] 

Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry, dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold.

Answers

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

Density = Mass / Volume

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Density = 301 g / 15.6 cm³Density = 19.3 g/cm³

The information below describes a redox reaction.
Ag+ (aq) + Al(s) — Ag(s) + Al3+ (aq)
Ag+ (aq) + -> Ag(s)
Al(s)->A3+ (aq) + 3e-
What is the coefficient of silver in the final, balanced equation for this reaction?

Answers

Answer:

Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

Explanation:

Oxidation:                            Al°(s) =>   Al⁺³(aq) + 3e⁻

Reduction:           3Ag⁺(aq) + 3e⁻ => 3Ag°(s)

_________________________________________

Net Rxn:           Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

One mole of neutral aluminum atoms (Al°(s)) undergo oxidation delivering 3 moles  of electrons to 3 moles silver ions (3Ag⁺³(aq)) that are reduced to 3 moles of neutral silver atoms (3Ag°(s)) in basic standard state 25°C; 1atm.

A balanced equation obeys the law of conservation of mass. According to the law of conservation of mass, mass can neither be created nor be destroyed. The coefficient of silver is 3.

What is a balanced equation?

A balanced chemical equation can be defined as the chemical equation in which the number of reactants and products on both sides of the equation are equal. The amount of reactants and products on both sides of the equation will be equal in a balanced chemical equation.

The numbers which are used to balance the chemical equation are called the coefficients. The coefficients are the numbers which are added in front of the formula.

The balanced chemical equation for the given redox reaction is given as:

Al (s) + 3 Ag⁺ (aq)  → Al³⁺ (aq) + 3Ag (s)

Thus the coefficient of silver is 3.

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Chrysanthenone is an unsaturated ketone. If Chrysanthenone has M+ = 150 and contains 2 double bond(s) and 2 ring(s); what is its molecular formula? Enter the formula in the form CH first, then all other atoms in alphabetical order; do not use subscripts. The formula is case-sensitive.

Answers

Answer:

the Molecular formula will be; C10H14O

Explanation:

Given the data in the question;

Chrysanthenone is an unsaturated ketone,

it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).

molecular formula = ?

we know that ketone contain 1 oxygen and mass of oxygen is 16

so mass of the C and H remaining will be;

⇒ 150 - 16 = 134

Now we determine the number of C atoms;

⇒ 134 / 13 = 10

hydrocarbon with 10 hydrogen atom have CnH2n+2 means

⇒ ( 10 × 2 ) +2 = 22 hydrogens

But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens

⇒ 22 - 6 - 2 = 14

Hence the Molecular formula will be; C10H14O

chemistry please help​

Answers

a. Process: I think it’s the Haber Process.

b. The substance represented by the letter C is Hydrogen.

c. The conditions represented by Y and Z needed for the process are 450 degrees Celsius and 200 atm, respectively.

d. i. The catalyst used is a finely divided iron (Fe) catalyst.
ii. I think the main ore from which the catalyst is extracted is Iron.

e. Balanced chemical equation for the reaction shown in the diagram:
N2(g) + 3H2(g) ⇆ 2NH3 (g)

f. One use of Ammonia is as a fertilizer.

I WILL GIVE BRAINLIEST IF YOU ANSWER CORRECTLY


How many grams in 3.75 x 1024 atoms of F?​

Answers

Answer:

96 grams

Explanation:

What causes an ice cube to melt when removed from a freezer?

Answers

Answer:

the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer

The temperature in the room is higher than the temperature in the freezer, thus causing the melting state of the ice cube when it is removed from the freezer.

Which is an example of poor safety practices when working outdoors

Answers

Answer:

C

Explanation:

u can't touch a chemical with bare skin

Answer:

touching a chemical with his or her bare skin.

To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7 mL of distilled water. Determine the molal concentration (m) of the resulting solution. MWLiCl

Answers

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

The molal concentration of the resulting solution is 0.492 m. The concentration number of moles of solute in one kg of solvent.

What is Molal concentration?

It is the measure of the concentration number of moles of solute in one kg of solvent.

To calculate the molal concentration first calculated the mass of [tex]\bold {LiCl}[/tex],

[tex]\rm \ mass = density \times volume[/tex]

Put the values,

m = 1 g/mL x  19.7 mL

m = 19.7 g     or

m = 0.0197 kg

Calculate the moles of LiCl:

[tex]n =\rm \dfrac { 0.411 \ g \times 1 mol }{ 42.394 \ g }\\\\n = 9.69\times 10^{-3} \ moles[/tex]

So, now the molal concentration,

[tex]m = \rm \dfrac { 9.69\times 10^{-3}\ mol} {0.0197 \ kg} \\\\m = 0.492[/tex]

Therefore, the molal concentration of the resulting solution is 0.492 m.

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Calculate the molality of each of the following solutions:
(a) 14.3 g of sucrose (C12H22O11) in 685 g of water,
(b) 7.15 moles of ethylene glycol (C2H6O2) in 3505 g of water.

Answers

Answer:

(a) The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]

(b)The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]

Explanation:

The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

Molality is expressed in units [tex]\frac{moles}{kg}[/tex]

(a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 [tex]\frac{g}{mole}[/tex], then 14.3 grams of the compound represents the following number of moles:

[tex]14.3 grams*\frac{1 mole}{342 grams} =[/tex] 0.042 moles

Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:

[tex]molality=\frac{0.042 moles}{0.685 kg}[/tex]

Solving:

molality= 0.0613[tex]\frac{moles}{kg}[/tex]

The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]

(b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:

[tex]molality=\frac{7.15 moles}{3.505 kg}[/tex]

Solving:

molality= 2.04[tex]\frac{moles}{kg}[/tex]

The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]

Write a word equation for
H2O2 + liver

Answers

Answer:

h2o2 and liver produce oxygen gas

Why is the reaction between potassium (K) and chlorine (Cl) considered an oxidation-reduction reaction? (4 points) K(s) + Cl2(g) → 2KCl(s) Select one: a. Chlorine reduces in size when it reacts with potassium. b. Chlorine takes electrons from potassium in the reaction. c. Potassium reduces in strength when it bonds with chlorine. d. Potassium takes chlorine's place in the reaction.

Answers

Answer:

(b) Chlorine takes electrons from potassium in the reaction.

Explanation:

Oxidation-Reduction: A reaction in which electrons are exchanged from one substance to another, also called redox.

In K(s) + Cl2(g) → 2KCl(s) , the electrons are being exchanged between one another. In this case, Chlorine atoms are taking electrons from potassium.

you're welcome

The reaction has been termed the oxidation-reduction reaction, as Chlorine takes electrons from potassium. Thus, option C is correct.

The reaction between Chlorine and Potassium has resulted in the formation of Potassium Chloride. The reaction has been termed the ionic redox reaction.

The redox reaction can be described as the oxidation-reduction reaction in which the oxidation of an atom by the loss of electrons, results in the reduction of another atom by the gain of the electron.

In the reaction of Chlorine and Potassium, there has been the loss of electron potassium and the gain of electrons by Chlorine. Thereby Potassium is getting oxidized, and Chlorine is getting reduced.

Thus, the reaction has been termed the oxidation-reduction reaction, as Chlorine takes electrons from potassium. Thus, option C is correct.

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HELPP!
how many grams are in 1.50 x 10(24) atoms of copper?

Answers

Answer:

2266g

Explanation:

mass = no.of molecules /6.o23*1o(23) * molar mass

molar mass of co2= 44g /mol

1.5 .10^25/6.023 .10^23 =51.5 moles of co2

51.5 .44g/mol =2266 g

1 mole is equal to 6.023 × 10 ²³ molecules. 2266 grams are in 1.50 x 10²⁴ atoms of copper.

What do you mean by mole ?

The term mole is defined as the amount of substance of a system which contains as many elementary entities.

One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.

The mole concept can be used to convert between mass and number of particles.

mass = number of molecules / 6.023 × 10²³ × molar mass

molar mass of CO2 = 44g /mol

1.5 .10^25 / 6.023 × 10^23

= 51.5 moles of CO2

= 51.5 .44g/mol

= 2266 g

Thus, 2266 grams are in 1.50 x 10²⁴ atoms of copper.

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If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give your reason.​

Answers

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Which must be balanced according to the law of conservation of mass:

[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]

Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.

Regards!

Three acid samples are prepared for titration by 0.01 M NaOH:

1. Sample 1 is prepared by dissolving 0.01 mol of HCl in 50 mL of water.
2. Sample 2 is prepared by dissolving 0.01 mol of HCl in 60 mL of water.
3. Sample 3 is prepared by dissolving 0.01 mol of HCl in 70 mL of water.

a. Without performing a formal calculation, compare the concentrations of the three acid samples (rank them from highest to lowest).
b. When the titration is performed, which sample, if any, will require the largest volume of the 0.01 M NaOH for neutralization?

Answers

B is the correct answer

The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?

2C6H6 (l) + 15O2 (g) → 12CO2 (g) + 6H2O (l)

Answers

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

For the given reaction, 2 moles of C₆H₆ the heat energy released is - 6535 KJ. Then, for 16 g of the compound or 0.205 moles needs 669.83 KJ of heat released in combustion.

What is combustion ?

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, typically oxygen, resulting in the release of heat, light, and various combustion products, such as carbon dioxide and water vapor.

The process of combustion involves a rapid and exothermic (heat-releasing) oxidation reaction that produces a flame, which is visible in many cases.

Here, 2 moles of the hydrocarbon releases - 6535 KJ of energy.

molar mass of C₆H₆ = 78 g/mol

then no.of moles in 16 g = 16 /78 = 0.205 moles.

then energy released by 0.205 moles  = 0.205 moles × 6535 KJ /2 moles = 669.83 kJ

Therefore, the heat energy released by 16 g of the compound in combustion is 669.83 kJ.

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Plutonium-238 is a radioactive element that decays over time into a less harmful element at a rate of 0.8 % per year. A power plant has 50 pounds of plutonium-238 to dispose of. How much plutonium-238 will be left after 10 years

Answers

Answer:

46 pounds

Explanation:

Let the amount of Plutonium-238 left after 10 years be P

Let the initial mass of Plutonium-238 be Po

Let the rate of decay be r

Let the time taken be t

Hence;

P = Poe^-rt

P= 50e^-0.008×10

P= 46 pounds

What is the pOH of 0.5 M KOH?

Answers

Answer:

pOH = 0.3

Explanation:

As KOH is a strong base, the molar concentration of OH⁻ is equal to the molar concentration of the solution. That means that in this case:

[OH⁻] = 0.5 M

With that information in mind we can calculate the pOH by using the following formula:

pOH = -log[OH⁻]pOH = -log(0.5)pOH = 0.3

QUESTIONS :
1.
Many of the flavours and smells of fruits are esters. A learner prepared an ester with a sme
Ilke banana in the school laboratory using pentanol and ethanoic acid. She set up the
apparatus as shown in the diagram below.
PAPER TOWEL DIPPED
-WATER BATH
IN COLD WATER
PEITANOL ETHANOIC
ACID+ 4 DROPS OF
SULPHURIC ACID
1.1 Which property of sulphuric acid makes it suitable to use as a catalyst for the
preparation of esters?
1.2 Why do we heat the test tube in a water bath and not directly over a flame?
1.3 With reference to the characteristic smells of esters, name TWO examples where
esters are used in different industries.
1.4 State ONE function of the wet paper towel in the opening of the test tube.
1.5 Write down the IUPAC name of an ester fomed.​

Answers

Answer:

See explanation

Explanation:

Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.

Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.

The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.

Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.

The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.

The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.

cấu hình electron của nguyên tử Ca

Answers

Explanation:

Do đó cấu hình electron của canxi là: 1s2 2s2 2p6 3s2 3p6 4s2.

What happens to an electroscope when a negatively charged rod is brought close to the metal sphere at the top?

Answers

Answer:

When the negatively-charged rod is brought close to the electroscope, positive charges are attracted to it and negative charges are repelled away from it. The electroscope has a net neutral charge and the rubber rod has a net negative charge. If they are brought into contact, they will both take a net negative charge.

Explanation:

I looked it up

Guanidine is a neutral compound but is an extremely powerful base. In fact, it is almost as strong a base as a hydroxide ion. Identify which nitrogen atom in guanidine is so basic, and explain why guanidine is a much stronger base than most other amines.

Answers

Answer:

See explanation

Explanation:

The molecule called guanidine is shown in the image attached to this answer. It contains three nitrogen atoms. Two among them are sp3 hybridized while one of them is sp2 hybridized.

Guanidine is more basic than other amines because its protonanation leads to the formation of three equivalent resonance structures thereby making its protonated form quite stable. This effect is not observed in other amines.

Also, the sp2 hybridized nitrogen atom is more basic and more easily protonated because when it is protonated, three equivalent resonance structures are obtained.

can yall please help im very slow

Answers

Answer:

turtle

Explanation:

they are slow and they take there time

What is used to measure P? To measure T? in Gas laws​

Answers

Answer:

P = pressure T = temperature

A sample of oxygen gas occupies a volume of 2.,0cm3 at pressure of 700K pa. what will be pressure of the same sample occupies a volume of 150cm, assume temperature remains constant​

Answers

Answer:

The pressure will be 933.33 Kpa

Explanation:

Given that:

Volume V₁ = 200 cm³  (note, there is a mistake in the volume. It is supposed to be 200 cm³)

Pressure P₁ = 700 Kpa

Pressure P₂ = ??? (unknown)

Volume V₂ = 150 cm³

Temperature = constant

Using Boyle's law:

PV = constant

i.e.

P₁V₁ = P₂V₂

700 Kpa × 200 cm³ = P₂ × 150 cm³

P₂ = (700 Kpa × 200 cm³)/150 cm³

P₂ = 933.33 Kpa

How many grams of KCl solid do I need to make 500 ml of 8% KCl solution

Answers

Answer:

40 grams of KCI are required to make 500 ml of 8% solution.

during the process of photosynthesis, green plants produce...

Answers

Answer: photosynthesis

Explanation:

they use carbon dioxide and energy from the sun to make sugar molecules and oxygen

A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
B) The rate of effusion of Xe gas through a porous barrier is observed to be 7.03×10-4 mol / h. Under the same conditions, the rate of effusion of SO2 gas would be ______ mol / h

Answers

Answer:

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

[tex]V_{gas}=\sqrt{\frac{8RT}{\pi M}}[/tex]

     OR

[tex]V_{gas}\propto \sqrt{\frac{1}{M}}[/tex]

where, M is the molar mass of gas

Forming an equation for the two gases:

[tex]\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}[/tex]          .....(1)

Given values:

[tex]V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol[/tex]

Plugging values in equation 1:

[tex]\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s[/tex]

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

[tex]Rate\propto \frac{1}{\sqrt{M}}[/tex]

Where, M is the molar mass of the gas

Forming an equation for the two gases:

[tex]\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}[/tex]          .....(2)

Given values:

[tex]Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol[/tex]

Plugging values in equation 2:

[tex]\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr[/tex]

Hence, the rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

An unknown organic compound composed of carbon, hydrogen and oxygen was analyzed and found to be 50.84% C, 8.53% H and the rest being oxygen. Which of the following represents the correct empirical formula for the compound?
(a) CH2O
(b) C3H602
(c) C4H803
(d) C2H40
(e) C5H1003

Answers

Answer:

e) C5H10O3

Explanation:

According to the information given in this question;

50.84% represents C

8.53% represents H

[100% - (50.84 + 8.53)]

= 100 - 59.37

= 40.63% of Oxygen (O)

This percentage means that;

Carbon = 50.84g

Hydrogen = 8.53g

Oxygen = 40.63g

By dividing by their respective atomic masses, we convert to moles:

C = 50.84g ÷ 12 = 4.24mol

H = 8.53g ÷ 1 = 8.53mol

O = 40.63g ÷ 16 = 2.54mol

Next we divide by the smallest no of the values (2.54mol)

C = 4.24 ÷ 2.54 = 1.669

H = 8.53 ÷ 2.54 = 3.358

O = 2.54 ÷ 2.54 = 1

To get a simple whole number ratio, we multiply the results by 3

C = 1.669 × 3 = 5.007

H = 3.358 × 3 = 10.074

O = 1 × 3 = 3

Simple whole number ratio of carbon, hydrogen and oxygen is 5:10:3. Hence, the empirical formula is C5H10O3.

You have reacted 100.00mL of 1.353M aqueous sulfuric acid with 12.618g of sodium hydroxide solid. If all of the heat generated by this reaction is transferred to a 1.317kg block of copper metal initially at 16.82°C, what is the final temperature of the block of copper metal? (Specific heat of copper = 0.375J/g*°C)

Answers

Answer:

70.137 °C

Explanation:

The reaction generated from the question can be expressed as:

[tex]2NaOH_{(s)} + H_2SO_{4(aq)} \to Na_2SO_{4(aq)} +2H_2O_{(l)}[/tex]

The enthalpy reaction:

[tex]\Delta H^0 _{rxn} = \sum \Delta H^0 _{f (products)} - \sum \Delta H^0 _{f (reactants)}[/tex]

[tex]\Delta H^0 _{rxn} =[2 \times \Delta H^0 _f (H_2O) + \Delta H^0 _f (Na_2SO_4) ] -[2 \Delta \times H^0 _f (NaOH) + \Delta H^0 _f (H_2SO_4) ][/tex]

Repacing the values of each compound at standard enthalpy conditions;

[tex]\Delta H^0 _{rxn} =[2 \times -279.4 + (-1384.49)]-[(2\times -418) -913]\ kJ[/tex]

[tex]\Delta H^0 _{rxn} =-194.29 \ kJ[/tex]

no of moles of NaOH = 12.618g/39.99 g/mol

= 0.3155 mol

no of moles of H₂SO₄ = molarity of H₂SO₄ × Volume

= 1.3553 mol/L × 100 × 10⁻³ L

= 0.13553 mol

From the reaction,

1 mol of NaOH = 2 × mol of H₂SO₄

Since mol of NaOH is greater than that of  H₂SO₄, then NaOH is the excess reagent and H₂SO₄ is the limiting reactant

1 mol of H₂SO₄ yields = - 194.29 kJ

0.13553 mol of H₂SO₄ will yield;

[tex]= \dfrac{-194.29 \ kJ}{1 \ mol} \times 0.13553 \ mol[/tex]

= -26.332124 kJ

= -26332.12 J

Finally,

Heat(q) = [tex]m_{(copper)} \times C_{(copper)}\times \Delta T[/tex]

26332.12 J = 1.317 × 10³ g × 0.375 J/g°C ×ΔT

26332.12 J = 493.875 J/° C  × ΔT

26332.12 /  493.875 = ΔT

ΔT = 53.317 °C

[tex]T_f - T_i = 53.317 ^0 C[/tex]

[tex]T_f[/tex]- 16.82 °C = 53.317 °C

[tex]T_f[/tex] = (53.317 + 16.82) °C

[tex]T_f[/tex] = 70.137 °C

Other Questions
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