Of the following substances, an aqueous solution of ________ will form basic solutions. NH4Br Pb(NO3)2 K2CO3 NaF

Answers

Answer 1

Answer:

K2CO3  and NaF

Explanation:

In order to ascertain which salt would form a basic solution we have to identify the classification of each of the salts.

- NH4Br: is the salt of a weak base (NH3) and a strong acid (HBr). This means that it would form an acidic solution.

- Pb(NO3): This is a normal salt, hence would not form a basic solution.

-  K2CO3: This is salt that forms a strongly alkaline/basic solution.

- NaF: it is the salt of a strong base, NaOH, and a weak acid, HF. This means this would form a basic solution.

Answer 2

The compounds capable to form basic solutions are[tex]\rm \bold {K_2CO_3 }[/tex] and NaF. Thus, options C and D are correct.

The basic solution has been given with the presence of a high number of hydroxide ions, while the acidic solution has been the presence of hydrogen ions.

The solution has been considered as basic when the compound has been constituted of a strong base. The constituents of the following compounds have been:

Ammonium bromide: The basic part is ammonia, and is a weak base. Thus, forms an acidic solution

Lead nitrate: The compound is salt and results in a neutral solution.

Potassium carbonate: The base has been carbonate, and a strong base. Thus forms the basic solution.

Sodium fluoride: The fluoride has been the basic part and has been a constituent of a strong base. It has been capable of forming a basic solution.

The compounds capable to form basic solutions are[tex]\rm \bold {K_2CO_3 }[/tex] and NaF. Thus, options C and D are correct.

For more information about the basic solution, refer to the link:

https://brainly.com/question/3595168


Related Questions

Describe the similarities between H3O and NH3. Compare/contrast their shapes and polarities within the context of your answer. These molecules are called isoelectronic. Why

Answers

Answer:

Explanation:

[tex]H_3O^+[/tex] also known as hydronium ion is formed as a result of the reaction between an hydrogen proton and a water molecules.

i.e [tex]\mathtt{H^+ + H_2O \to H_3O^+}[/tex]

(molecular geometry for the hydronium ion shows that the lewis structure of hydronium ion possess a three hydrogen ion bonded to a central atom known as oxygen. The oxygen possess a lone pair with a positive ion. So we have three hydrogen atoms and a lone pair attached to the oxygen. We can now say that there are four groups as the steric number in which one of them is a lone pair. This give rise to the trigonal pyramidal shape of the [tex]H_3O^+[/tex] (hydronium ion) with a bond angle of about 109,5°

Similarly, [tex]NH_3[/tex] on the other hand also known as ammonia has a shape that can be also determined by the Lewis structure.

IN ammonia,  there are three hydrogen  and a lone pairs of electron spreading out as far away from each other  from the centre nitrogen. In essence, the valence shell electron pair around hydrogens tend to repel each other. Hence, giving it a trigonal pyramidal shape.

From above the similarities between H3O and NH3 is in their molecular geometry in which both  H3O and NH3 have the same shape.

These molecules are called isoelectronic. Why?

Isoelectronic molecules are molecules having the same number of electrons and same electronic configuration  structure. As a result H3O and NH3 possess the same  number of electrons in the same orbitals and they also posses the same structure.

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which side of the reaction do they appear?

MnO41- (aq) + Cl1- (aq) → Mn2+ (aq) + Cl2 (g)

a. 2 moles of H2O on the reactant side
b. 2 moles of H2O on the product side
c. 4 moles of H2O on the product side
d. 8 moles of H2O on the product side
e. 10 moles of H2O on the reactant side

Answers

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

[tex]MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\[/tex]

Then, we add the half reactions:

[tex]2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0[/tex]

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

Experiment:
Part I: Voltaic Cell
Assume that you are provided with the following materials:
Strips of metallic zinc, metallic copper, metallic iron
1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine (I2)
Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos, identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.
For each cell created, include the following details.
Which electrode was the anode and which was the Cathode?
The anode and cathode half reactions.
Balanced equation for each cell you propose to construct.
Calculated Eocell
Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed)

Answers

Answer:

Here are four possible voltaic cells.  

Explanation:

1. Standard reduction potentials

                                         E°/V

I₂(s) + 2e⁻ ⟶ 2I⁻(aq);        0.54

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);   0.34

Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);   -0.41

Zn²⁺(aq) + 2e⁻ ⟶ Zn(s);   -0.76

2. Possible Voltaic cells

(a) Zn/I₂

                                                                       E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                 0.76

Cathode:  I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 1.30

Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Zn is the anode; graphite is the cathode.

(b) Zn/Cu²⁺

                                                                          E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                    0.76

Cathode:  Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);                  0.34

Cell:          Zn(s) +  Cu²⁺(s) ⟶  Zn²⁺(aq) + Cu(s); 1.10

Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)

Zn is the anode; Cu is the cathode.

(c) Zn/Fe²⁺

                                                                            E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                     0.76

Cathode:  Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);                    -0.41

Cell:          Zn(s) +  Fe²⁺(s) ⟶  Zn²⁺(aq) + Fe(s);  0.35

Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)

Zn is the anode; Fe is the cathode.

(d) Fe/I₂

                                                                         E°/V

Anode:     Fe(s) ⟶ Fe²⁺(aq) + 2e⁻;                   0.41

Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 0.95

Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Fe is the anode; graphite is the cathode.

 

Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced:________
A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol

Answers

Answer:

B. 1-amino-2-methylpentan-2-ol

Explanation:

In this case, the first step, we have the attack of the nucleophile cyanide ([tex] CN^-[/tex] produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.

Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas ([tex]H_2[/tex]) in the presence of a catalyst ([tex]Pd[/tex]), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.

See figure 1 to further explanations

I hope it helps!

If the rate of formation (also called rate of production) of compound C is 2M/s in the reaction A --->2C, what is the rate of consumption of A

Answers

Answer:

[tex]r_A=-1\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:

[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]

In such a way, solving the rate of consumption of A, we obtain:

[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]

Clearly, such rate is negative which account for consumption process.

Regards.

What is the primary source of energy in most living communities?

Answers

Answer:

The sun

Explanation:

The sun is the primary source of energy in most living communities. The producers or the green plants that prepare their own food by the use of sunlight and other natural resources. Carbon dioxide, water, and other minerals are used by the plants to make their food in the presence of chlorophyll. Plants are then consumed by the consumers. This chain helps in forming the food chain and the food web.

A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?

Answers

Explanation:

To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;

Step 1: Divide the percentage composition by the atomic mass

Sulphur = 31.42 / 32 = 0.9819

Oxygen = 31.35 / 16 = 1.9594

Flourine = 37.23 / 19 = 1.9595

Step 2: Divide by the lowest number

Sulphur = 0.9819 / 0.9819 = 1

Oxygen = 1.9594 / 0.9819 ≈ 2

Flourine = 1.9595 / 0.9819 ≈ 2

This means the ratio of the elements is 1 : 2: 2

The empirical formular (simplest formular of a compound) of the compound is;

SO₂F₂

To obtain the molecular formular (Actual formular of a compound);

(SO₂F₂)n = 102.1

Inserting the atomic masses and solving for n;

(102)n = 102.1

n ≈ 1

The molecular formular is; (SO₂F₂)₁ = SO₂F₂

I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.

Answers

Answer:

THE FINAL TEMPERATURE OF WATER IS -4.117 °C

Explanation:

Mass of the aluminium = 50 g

c = 0.88 J/g C

Initial temperature of aluminium = 225 °C

Volume of water = 100 ml

Density of water = 1 g/ml

Mass of water = density * volume of water

Mass of water = 1 * 100 = 100 g of water

Initial temperature of water = 20 C

It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,

So therefore,

heat lost by aluminium = heat gained by water

H = mass * specific heat capacity * temeprature change

So:

m c ( T2- T1) = m c (T2-T1)

50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)

44 ( T2 - 225 ) = 418 ( T2 - 20)

44 T2 - 9900 = 418 T2 - 8360

-9900 + 8360 = 418 T2 - 44 T2

-1540 = 374 T2

T2 = - 4.117

So therefore the final temperature of water is -4.117 °C

When the equation MnO₄⁻ + I⁻ + H₂O → MnO₂ + IO₃⁻ is balanced in basic solution, what is the smallest whole-number coefficient for OH⁻?

Answers

Answer:

The smallest whole-number coefficient for OH⁻ is 2

Explanation:

Step 1: The equation redox reaction is divided into two half equations

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 5 : addition of the two half equations to yield a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole number coefficient for OH⁻ is 2

A redox reaction is divided into two half equations which are shown below:

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The two half equations are then added and written together to form a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole-number coefficient for OH⁻ is therefore 2.

Read more on https://brainly.com/question/17156617

How many molecules are there in 3.5 moles of carbon dioxide? A. 63.21 x 10^23 B. 21.07 x 10^23 C. 42.14 x 10^23 D. 6.02 x 10^23

Answers

Answer:

B. 21.07 x 10²³ molecules

Explanation:

Avogadro's Number: 6.022 x 10²³

Step 1: Set up equation

[tex]3.5 mols CO_2(\frac{6.022(10^{23}) moleculesCO_2}{1 mol CO_2})[/tex]

Step 2: Multiply and cancel out units

3.5(6.022 x 10²³) = 21.07 x 10²³ molecules CO₂

Step 3: Convert to proper scientific notation

≈ 2.11 x 10²³ molecules CO₂

Select True or False: Pi bonds are covalent bonds in which the electron density is concentrated above and below a plane containing the nuclei of the bonding atoms and occurs by sideways overlap of p orbitals.

Answers

Answer:

True

Explanation:

In pi bonds, the electron density concentrates itself between the atoms of the compound but are present on either side of the line joining the atoms. Electron density is found above and below the plane of the line joining the internuclear axis of the two atoms involved in the bond.

Pi bonds usually occur by sideways overlap of atomic orbitals and this leads to both double and triple bonds.

explain how the liquid in a thermometer changes so that it can be used to measure a temprature

Answers

Answer:

The liquid that is often used in thermometers is chrome.

It is khwon for raising its volule when the temperature raises and vice-versa. ● the temperature and the volume are proprtional to each other so using Mathematics, scientists have figured out a way to benefit from it to make a thermometer.

In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?

a. Cyclohexanone
b. Ethyl Acetate
c. λ- butyrolactone
d. Pentanamide
e. Propanoyl Chloride

Answers

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the cyclohexanone.

See figure 1

I hope it helps!

In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270.

Use the information she obtained to determine the Ka for this acid.

Ka(experiment) = _____

Answers

Answer:

Ka = 6.87x10⁻⁵

Explanation:

The equilibrium of benzoic acid in water is:

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The equilibrium constant, Ka, is:

Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]

The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.

X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.

Concentrations in equilibrium are:

[C₆H₅COOH] = 0.425M - X[C₆H₅COO⁻] = X [H₃O⁺] = X

pH is defined as -log [H₃O⁺]. As pH = 2.270

2.270 = -log [H₃O⁺]

10^-2.270 = [H₃O⁺]

5.37x10⁻³M = [H₃O⁺] = X.

Replacing, concentrations in equilibrium are:

[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M

[C₆H₅COO⁻] = 5.37x10⁻³M

[H₃O⁺] = 5.37x10⁻³M

Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]

Ka = 6.87x10⁻⁵

PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?

Answers

Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]

So, the value of n is 0.207 mol.

A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell

Answers

Answer:

The answer is D) Electrolytic cell

Explanation:

An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.

When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.

These cells are the closest thing to a galvanic battery.

Answer:

b. voltaic cell

Explanation:

Founders Education answer. had to take this quiz 4 times

Refer to the figure.
30. How many planes are shown in the figure?
31. How many planes contain points B, C, and E?
32. Name three collinear points.
3. Where could you add point G on plane N
so that A, B, and G would be collinear?
4. Name a point that is not coplanar with
A, B, and C.
5. Name four points that are coplanar.
BN

Answers

Answer:

  30. 5 planes are shown

  31. 1 plane

  32. CEF

  33. on line AB

  34. E or F

   35. ABCD or BCEF or CDEF or ACEF

Explanation:

30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.

__

31. 3 points define one plane only.

__

32. The only points shown on the same line segment are points E, F, and C.

__

33. If G is to be collinear with A and B, it must lie on line AB.

__

34. The only points shown that are not on plane N are points E and F. Either of those will do.

__

35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...

ABCDBCEFCDEF

Plane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)

How did Jesseca Kusher create her new material?

Answers

Answer:

Jesseca Kusher, an 18-year-old researcher from Spartansburg, S.C., invented a paint-on coating for roofing shingles. Her formula could reduce a home's cooling costs and possibly cut ozone pollution in urban areas...

SUPPORT ME ...........

Answer:

Jesseca created mixtures containing graphite, gypsum, and mica that could be painted on roof shingles.

Explanation:

Hope this helped!!

What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)

Answers

Answer:

5.90

Explanation:

Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol

Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol

CH3COO- + HCl => CH3COOH + Cl-

Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol

Moles of CH3COOH formed = moles of HCl added = 0.0005 mol

pH = pKa + log([CH3COO-]/[CH3COOH])

= -log Ka + log(moles of CH3COO-/moles of CH3COOH)

= -log(1.78 x 10^(-5)) + log(0.007/0.0005)

= 5.90

Answer:

The correct answer is 5.895.

Explanation:

The reaction will be,

CHCOO⁻ + H+ ⇔ CH₃COOH

Both the HCl and the acetate are having one n factor.

The millimoles of CH₃COO⁻ is,

= Volume in ml × molarity = 10 × 0.75 = 7.5

The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5

Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0

The volume of the solution is, 10+5 = 15 ml

The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15

The molarity of CH₃COOH is 0.5/15

pH = pKa + log[CH₃COO⁻]/[CH₃COOH]

= 4.74957 + 1.146

= 5.895

When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l)
A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%

Answers

Answer:

D) 65.7%

Explanation:

Based on the reaction:

2H2(g)+O2(g)⟶2H2O(l)

2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.

To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.

Theoretical yield:

Moles of 5.58g H₂:

5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂

As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:

2.768 moles H₂O ₓ (18.015g / mol) =

49.86g H₂O is theoretical yield

Percent yield:

Percent yield = Actual yield / Theoretical yield ₓ 100

32.8g H₂O / 49.86g ₓ 100 =

65.7% is percent yield of the reaction

D) 65.7%

Assume that 33.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.How many moles of have been added at the equivalence point?n = ? mol

Answers

The answer would have to be 33 moles

Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the same, 10 B. the same, 14 C. a different, 6 D. a different, 10

Answers

Answer:

Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons

Explanation:

The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.

When these electrons change orbit or level they release energy in the form of light, which is known as a photon.

Which of the following is not the same as 1,400 mL? a. 1.4 cm³ b 1.4 L c. 1,400 cm³ d. 140 cL

Answers

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:

A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.

Answers

Answer:

Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.

Explanation:

Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.

Small amounts of calcium hydroxide salt,  [tex]Ca(OH)_{2}_(s)[/tex]  is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.

I hope this explanation is helpful.

Assume you dissolve 0.235 g of the weak benzoic acid, C6H5CO2H in enough water to make 100.0 mL of the solution and then titrate the solution with 0.108 M NaOH. Benzoic acid is a monoprotic acid.
1. What is the pH of the original benzoic acid solution before the titration is started?
2. What is the pH when 7.00 mL of the base is added? (Hint: This is in the buffer region.)
3. What is the pH at the equivalence point?

Answers

Answer:

1. pH = 2.98

2. pH = 4.02

3. pH = 8.12

Explanation:

1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:

0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M

The equilibrium of benzoic acid with water is:

C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)

And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:

Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]

The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:

[C6H5CO2H] = 0.01924 - X

[C6H5O⁻] = X

[H3O⁺] = X

Replacing in Ka:

6.14x10⁻⁵ = [X] [X] / [0.01924 - X]

1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²

1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0

Solving for X:

X = -0.0010→ False solution. There is no negative concentrations

X = 0.0010567M → Right solution.

pH = - log [H3O⁺] and as [H3O⁺] = X:

pH = - log [0.0010567M]

pH = 2.98

2.

pH of a buffer is determined using H-H equation (For benzoic acid:

pH = pka + log [C6H5O⁻] / [C6H5OH]

pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical

The benzoic acid reacts with NaOH as follows:

C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O

That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)

7.00mL of NaOH 0.108M are:

7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻

And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH

Replacing in H-H equation:

pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]

pH = 4.02

3. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles

Volume of NaOH to reach equivalence point:

1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:

1.164x10⁻³ moles / 0.111L = 0.01049M

Equilibrium of  C₆H₅O⁻ with water is:

C₆H₅O⁻(aq) + H₂O(l) ⇄  C₆H₅OH(aq) + OH⁻(aq)

Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]

Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰

Equilibrium concentrations of the species are:

C₆H₅O⁻ = 0.01049M - X

C₆H₅OH = X

OH⁻ = X

Replacing in Kb expression:

1.63x10⁻¹⁰ = X² / 0.01049- X

1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0

Solving for X:

X = -1.3x10⁻⁶ → False solution

X = 1.3076x10⁻⁶ → Right solution

[OH⁻] =  1.3076x10⁻⁶

as pOH = -log [OH⁻]

pOH = 5.88

And pH = 14 - pOH

pH = 8.12

What is the formula of a compound if a sample of the compound contains 0.492 mol X, 0.197 mol Y, and 0.295 mol Z?

Answers

Answer:

X₅Y₂Z₃

Explanation:

The formula of a compound is determined as the whole number ratio between moles of each element present in the molecule.

The molecule is made from X, Y and Z. To fin the ratio we will divide the given moles in the moles of Y (0.197 moles), because is the element with the low number of moles.

X = 0.492 moles / 0.197 moles = 2.5

Y = 0.197 moles / 0.197 moles = 1

Z = 0.295 moles / 0.197 moles = 1.5

But, as the formula is given just with whole numbers, if we multiply each number twice:

X = 2.5*2 = 5

Y = 1*2 = 2

Z = 1.5*2 = 3

The formula is:

X₅Y₂Z₃

I think I'm typing it into my calculator wrong. I will give brainliest to whoever gets it right.

Answers

Answer:

36.7 mg

Explanation:

The following data were obtained from the question.

Original amount (A₀) = 65.1 mg

Rate constant (K) = 2.47×10¯² years¯¹

Time (t) = 23.2 years

Amount of substance remaining (A) =?

Thus, we can obtain the amount of substance remaining after 23.2 years as follow

ln A = lnA₀ – Kt

lnA = ln(65.1) – (2.47×10¯² × 23.2)

lnA = 4.1759 – 0.57304

lnA = 3.60286

Take the inverse of ln

A = e^3.60286

A = 36.7 mg

Therefore, the amount remaining after 23.2 years is 36.7 mg.

Complete the sentences describing the cell.

a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.

Answers

Answer:

a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.

b. Therefore electrons flow from the aluminium electrode to the nickel electrode.

c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.

Explanation:

Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.

Given:

E ⁰N i ⁺² = − 0.23 V   is the standard reduction potential for the nickel ion

E ⁰  A l ⁺³ =  − 1.66  V  is the standard reduction potential for the aluminum ion

The most negative potentials  correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.

So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.

The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.

Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.

Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.

So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.

the ka of hypochlorous acid (hclo) is 3.0 x10^-8 at 25.0°C. What is the % of ionization of hypochlorous

Answers

Answer:

0.14%

Explanation:

The computation of % is shown below:

As we know that

     HClO <=> H+ + ClO-

I         0.015          0      0

C          -a            +a     +a

E      0.015-a        a       a

Now

[tex]Ka = \frac{[H+][ClO-]}{[HClO]}[/tex]

[tex]= \frac{a^{2}}{(0.015 - a)} \\\\= 3.0 \times 10^{-8}[/tex]

[tex]a^{2} + 3.0 \times 10^{-8}a - 4.5 \times 10^{-10} = 0[/tex]

Now Solves the quadratic equation i.e.

[tex]a = 2.120 \times 10^{-5}[/tex]

[tex][H+] = a = 2.120 \times 10^{-5} M[/tex]

So,

% ionization is

[tex]= \frac{[H+]}{[HClO]}_{initial} \times 100\%\\\\= 2.120 \times 10^{-5}\div0.015 \times 100\%[/tex]

= 0.14%

Hence, the percentage of hypochlorous ionization is 0.14%

At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with X Benzene

Answers

CHECK COMPLETE QUESTION BELOW

At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.

Answer:

The total vapor pressure is [tex]81.3 mmHg[/tex]

Explanation:

We will be making use of Dalton and Raoults equation in order to calculate the total pressure,

Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]

PT= total vapor pressure

From the question

Benene's Mole fraction = 0.580

then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.

= (1 - 0.580) = 0.420

Vapor pressure of benzene given = 183 mmHg

Vapor pressure of toluene given= 59.2 mmHg

If we substitute those value into above equation, we have

PT=(183×0.580)+(59.2×0.420)

=81.3mmHg

Therefore,, the total vapor pressure of the solution is 81.3 mmHg

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