Sean plated an unknown metal onto his silver ring which initially weighed 1.4 g. He constructs an electrolytic cell using his ring as one of the electrodes. After running the cell, 0.022 moles of the unknown metal was plated onto his ring and the mass of the ring increased to 3.137 g. What is the atomic weight of the unknown metal in g/mol

Answer:

0.696 m

Explanation:

We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:

Mass of C₆H₁₂O₆ = 16.7 g

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)

= 72 + 12 + 96

= 180 g/mol

Mole of C₆H₁₂O₆ =?

Mole = mass / molar mass

Mole of C₆H₁₂O₆ = 16.7 / 180

Mole of C₆H₁₂O₆ = 0.093 mole

Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

133.6 g = 133.6 g × 1 Kg / 1000 g

133.6 g = 0.1336 Kg

Thus, 133.6 g is equivalent to 0.1336 Kg.

Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:

Mole of C₆H₁₂O₆ = 0.093 mole

Mass of water = 0.1336 Kg

Molality =?

Molality = mole / mass of water (in Kg)

Molality = 0.093 / 0.1336

Molality = 0.696 m

Therefore, the molality of the solution is 0.696 m

Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.

Group of answer choices

hydrochloric acid

sodium bicarbonate

calcium carbonate

sodium chloride

sodium hydroxide

Answer:

calcium carbonate

Explanation:

A stalactite is an icicle-looking mould that is formed by the precipitation of natural minerals as a result of water dripping from the ceiling, hanging from a cave.

A stalagmites in the other hand, grows upwards and is also a mound that is formed by the deposits of minerals gotten by the water dripping on the floor of a cave.

Therefore, stalactites and stalagmites form as calcium carbonate precipitates out of the water evaporating in underground caves.

Answer:

0.0063 mol

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(g)

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of C₈H₁₈ to CO₂ is 1:8.

Step 3: Calculate the number of moles of C₈H₁₈ needed to produce 0.050 moles of CO₂

0.050 mol CO₂ × 1 mol C₈H₁₈/8 mol CO₂ = 0.0063 mol C₈H₁₈

Answer:

MoO2

Explanation:

The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.

The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239.9g/mol-

12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo

Mass Mo -95.95g/mol-:

0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo

Mass oxygen in the oxide:

13.197 - 9.895g = 3.302g Oxygen

Moles oxygen -Molar mass: 16g/mol-:

3.302g Oxygen * (1mol / 16g) = 0.206 moles O

Now, the ratio of moles O / moles Mo is:

0.206 moles O / 0.1031 moles Mo = 2

That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:

Answer:

i. HCIO2 - 2.9

ii. CH3COOH - 2.64

iii. HF - 3.27

iv. HNO2 - 2.67

Explanation:

The Ph value differ for base and acid. A neutral solution will have Ph value of 7. For acid the Ph value is less than 7 and for base Ph value is greater than 7. The highest Ph value for the given acids is 3.27 for hydroflouric acids.

The pH of the resulting solution is 3.16. Acetic acid having the formula [tex]CH_3COOH[/tex] would produce the highest pH at the equivalence point in a weak acid-strong base titration.

Kingsley adds 49.28 mL of NaOH to 250.00 mL of the HCOOH solution.

The neutralization reaction resulted in 0.098 moles of HCOOH and 0.025 moles of HCOO⁻ left in the solution.

We have to determine the pH of the resulting solution and identify the acid that would produce the highest pH at the equivalence point in a weak acid-strong base titration.

pH, a quantitative measure of the acidity or basicity of aqueous or other liquid solutions.

Kingsley adds 49.28 mL of NaOH to 250.00 mL of the HCOOH solution.

Total volume of the solution = 49.28 mL + 250.0 mL = 299.28 mL = 0.29928 L

At neutralization point, HCOOH = 0.098 mol = [tex]\frac{0.098\ mol}{0.29928\ L} = 0.32745\ M[/tex]

So, Concentration of HCOOH at neutralization point = 0.32745 M

And [tex]HCOO^-[/tex] at neutralization point = 0.025 mol = [tex]\frac{0.025\ mol}{0.29928\ L} = 0.08353\ M[/tex]

So, Concentration of HCOO⁻ at neutralization point = 0.08353 M

Now, [tex]K_a[/tex] value of HCOOH = [tex]1.77\times10^{-4}[/tex] (reference value}

We know that [tex]pK_a = -logK_a[/tex]

[tex]pK_a = -log(1.77\times10^{-4}) = 3.75[/tex]

Now, HCOOH  is acid and HCOO⁻ is the conjugate base.

According to Henderson Hasselblach equation:

[tex]pH = pK_a + log\frac{[conjugate\ base]}{[acid]}\\ \therefore pH = 3.75 + log\frac{[HCOO^-]}{[HCOOH]}\\pH = 3.75 + log(\frac{0.08353}{0.32745})\\pH = 3.75 +(-0.59)\\pH = 3.16[/tex]

Therefore, the pH of the resulting solution is 3.16.

When weak acid is titrated with a strong base, the pH at the neutralization point is always greater than 7. It is because the conjugate base produced by weak acid in neutralization reaction is comparatively stronger in nature which produces [tex]OH^-[/tex] at neutralization point after reacting with water. so the solution is basic. As a result, pH is more than 7.

Now, the weaker is the acid, the stronger is the conjugate base, and vice-versa. Therefore, the concentration of [tex]OH^-[/tex] ions is more at the neutralization point. So more will be the pH value.

Given acids with the corresponding [tex]K_a[/tex] value are shown below:

[tex]K_a(HF) =6.8\times10^{-4}\\K_a(HNO_2) =4.5\times10^{-4}\\K_a(HClO_2) =1.1\times10^{-2}\\K_a(CH_3COOH) =1.8\times10^{-5}[/tex]

We know that lower the value of [tex]K_a[/tex] of  acid, the weaker is the acid and vice-versa. Hence, [tex]CH_3COOH[/tex] has lowest [tex]K_a[/tex] value compared to other acids. Therefore, [tex]CH_3COOH[/tex]  would  produce the highest pH at the equivalence point in a weak acid-strong base titration.

Hence, The pH of the resulting solution is 3.16. Acetic acid having the formula [tex]CH_3COOH[/tex] would produce the highest pH at the equivalence point in a weak acid-strong base titration.

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Answer:

heat; temperature; surrounding; insulated.

Explanation:

A calorimeter can be defined as a scientific instrument (device) that is designed and developed for measurement of the heat involved in chemical reactions or other processes, especially by taking the measurement of the temperature of the materials surrounding the process.

Basically, a calorimeter is insulated using materials with a very high level of resistivity, so as to prevent heat transfer to the outside of the device (calorimeter).

Some of the components that make up a simple calorimeter are; thermometer, an interior styrofoam cup, an exterior styrofoam cup, cover, etc.

Additionally, a calorie refers to the amount of heat required to raise the temperature of a gram of water by one degree Celsius (°C)

A calorimeter measures the heat involved in reactions or other processes by measuring the temperature of the materials surrounding the process. The calorimeter is insulated to prevent transfer of heat to outside the device.

Calorimeter is an instrument used in chemical reactions for measuring the heat, physical changes as well as heat capacity of the chemical reaction.

From the definition of calorimeter it is clear that it checks the temperature of the reaction and calculates the heat of the chemical reaction. It is insulated by those materials which are bad conductors of heat so that no heat will loss from this apparatus.

Hence, correct options are heat, temperature, surroundings and insulated.

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Answer:

35g/l

Explanation:

Salinity is practically the saltiness of the water, in basic terms. It's basically the amount of salt dissolved in water. Ocean water has a salinity of around 35g/l (that's about 3.5% of which is dissolved salt in water!). However, Atlantic Ocean (the ocean with the saltiest water), can range up to 37g/l.

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Answer:

a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :

There are no insoluble precipitate forms.

b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :                        

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.

                                          KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex]  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.

                                        [tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex]  ⇒ soluble.

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c)

Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.

                                        [tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex]  ⇒ insoluble.

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d)

Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.

                                        [tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex]  ⇒ soluble.

There are no insoluble precipitates forms.

Answer:

See explanation

Explanation:

The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;

k= Ae^-Ea/RT

Where;

k= rate constant

A= pre-exponential factor

Ea=activation energy

R= gas constant

T= temperature

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.

Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.

The correct answer is mentioned below.

The equation is as follows:-

[tex]k= Ae^{-Ea/RT[/tex] Where;

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

Hence, the correct answer is mentioned above.

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Answer:

Explanation:

Assume we have 100g of this substance. That means we would have 20.24g of Cl and 79.76g of Al. Now we can find how many moles of each we have:

[tex]\frac{79.76 \:g}{35.45 \: g/mol}[/tex] = 2.25 mol of chlorine

[tex]\frac{20.24 \: g}{26.98 \: g/mol}[/tex] = 0.750 mol of Al.

To form a integer ratio, do 2.25/0.75 = 2.99999 ~= 3.

So the ratio is essentially Al : Cl => 1 : 3. To the compound is possibly [tex]AlCl_3[/tex].

However, it says it has a molar mass of 266.64 g/mol, and since AlCl3 has a molar mass of 133.32, it must be [tex]Al_2Cl_6[/tex].

Actually this molecule isn't exactly AlCl3 (which is ionic). Al2Cl6 forms a banana bond where Cl acts as a hapto-2 ligand. But that's a bit advanced. All you need to know is X = Al2Cl6

The molecular formula of the compound is Al₂Cl₆

To solve the question given above, we'll begin by obtaining the empirical formula of the compound. This can be obtained as follow:

Aluminum (Al) = 20.24%

Chlorine (Cl) = 79.76%

Al = 20.24%

Cl = 79.76%

Divide by their molar mass

Al = 20.24 / 27 = 0.75

Cl = 79.76 / 35.5 = 2.25

Divide by the smallest

Al = 0.75 / 0.75 = 1

Cl = 2.25 / 0.75 = 3

Thus, the empirical formula of the compound is AlCl₃

Finally, we shall determine the the molecular formula of the compound.

Molar mass of compound = 266.64 g/mol

Empirical formula = AlCl₃

[AlCl₃]ₙ = 266.64

[27 + (3×35.5)]n = 266.64

[27 + 106.5]n = 266.64

133.5n = 266.64

Divide both side by 133.5

n = 266.64 / 133.5

Molecular formula = [AlCl₃]ₙ

Molecular formula = [AlCl₃]₂

Therefore, the molecular formula of the compound is Al₂Cl₆

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Answer:

4.858 g

Explanation:

Start with the formula

density = [tex]\frac{mass}{volume}[/tex]

density = 1.98 g/mL

volume = 2.45 mL

mass = ??

rearrange the formula to solve for mass

(density) x (volume) = mass

Add in the substitutes and solve for mass

1.98 g/mL x 2.45 mL = 4.858 g

Answer:

Every atom of a given element has the same number of protons

The protons in the nucleus of the helium atom are held together by a force called the strong nuclear force.

Explanation:

Atoms are composed of electrons, protons and neutrons. The electron is negatively charged, protons are positively charged and the neutrons have no charge.

Electrons are found in shells while protons are found inside the atomic nucleus. Similar to electrostatic forces between electron and proton, protons of helium are held together by a strong nuclear blinding force.

Note that, all isotopes must have the same atomic number. This shows that they are all the same atom changed by differences in number of neutrons.

Answer:

[tex]K=1.7x10^{-3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:

[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]

Next, we plug in the given concentrations on the data table to obtain:

[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]

Regards!

Answer:

(B). it's metallic bonding

Answer:

An indicator added to the solution turns green-blue.

Explanation:

An indicator is a substance which shows the degree of acidity or alkalinity of a solution by change in colour of the solution.

The universal indicator changes colour as the pH of the solution changes. Looking at the colour ranges for the universal indicator, the green-blue colour indicates a weak base.

Answer:

62.5 ml of 0.032 M  KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄

Explanation:

The balanced reaction is:

2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]

In this case, 50 mL (0.05 L) of 0.1 M H₂C₂O₄ react. So, replacing the data in the definition of molarity:

[tex]0.1 M=\frac{number of moles of solute}{0.05 L}[/tex]

Solving:

number of moles of solute= 0.1 M*0.05 L

number of moles of solute= 0.005 moles

So, 0.005 moles of H₂C₂O₄ react.  Then you can apply the following rule of three: if by stoichiometry 5 moles of H₂C₂O₄ react with 2 moles of KMnO₄, 0.005 moles of H₂C₂O₄ react with how many moles of KMnO₄?

[tex]moles of KMnO_{4} =\frac{0.005moles of H_{2} C_{2} O_{4}* 2moles of KMnO_{4} }{5moles of H_{2} C_{2} O_{4} }[/tex]

moles of KMnO₄= 0.002 moles

Knowing that the molarity of KMnO₄ is 0.032 M, replacing in its definition and solving:

[tex]0.032 M=\frac{0.002 moles}{volume}[/tex]

[tex]volume=\frac{0.002 moles}{0.032 M}[/tex]

volume= 0.0625 L= 62.5 mL

62.5 ml of 0.032 M  KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄

Answer:

Explanation:

When Tollen's test is done by aldehyde , silver ion is converted into silver which forms a layer which looks like a mirror.

Ag⁺ + e = Ag

It is a reduction process where silver(1) ion is reduced to metallic silver.

Aldehyde is oxidized to carboxylate ion.

CH₃CHO + 2 OH⁻ = CH₃COO⁻ + H₂O + H⁺ + 2e

Visual indicator is silver metal which forms silver mirror at the bottom of test tube .

Answer:  inference because she drew a conclusion based on evidence.

why?:

Because the evidence was that she heard the crashing sound, and then when she came into her room saw the broken window and baseball.

It was not an observation because she did not directly see the baseball going through the window

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

Explanation:

here's the answer to your question

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

Answer:

77%

Explanation:

First we convert 3.0 g of cyclohexanol (C₆H₁₂O) to moles, using its molar mass:

Then we convert 1.9 g of cyclohexene (C₆H₁₀) to moles, using its molar mass:

Finally we calculate the theoretical yield:

Answer:

1. Nitrate ions, NaNO3 - Sodium nitrate.

2. Sulphide ions, K2S - Potassium sulphide.

3. Sulphate ions, CaSO4 - Calcium sulphate.

4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.

5. Carbonate ions, CaCO3 - Calcium carbonate.

6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.

7. Phosphite ions, PH3 - Hydrogen phosphite.

8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).

9. Ethanoate ions, CH3COONa - Sodium ethanoate.

10. Methanoate ions, HCOONa - Sodium methanoate.

11. Fluoride ions, HF - Hydrogen fluoride.

12. Chloride ions, KCl - Potassium chloride.

13. Bromide ions, HBr - Hydrogen bromide.

14. Iodide ions, NaI - Sodium iodide.

15. Phosphate ions, K3PO3 - potassium phosphate.

Answer:

A) Forms a racemic mixture of the two possible enantiomers

When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and  new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

There are three types of chemical changes:

1) inorganic changes

2)organic changes

3) biochemical changes

During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.

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Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

Answer:

Explanation:

maleic acid ⇒ fumaric acid

ΔHreaction = ΔHproduct - ΔHreactant

ΔHproduct = -1336.0 kJ mol⁻¹

ΔHreactant = - 1359.2 kJ mol⁻¹.

ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)

=   1359.2 kJ mol⁻¹   -1336.0 kJ mol⁻¹

= 23.2 kJ mol⁻¹ .

Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.

b) (√)

c)(✓)

hsjdhfjdkskkshd

Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

1.8×10⁻⁶ = x²

√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH