A cube with sides 3cm.Its mass is 54g. will it sink or float

Answer:

B

Explanation:

B

Answer:

a. pH → 8.12

b. pH → 10.55

c. pH → 8.38

Explanation:

In this excersise, the first two salts come from a strong base and a weak acid, so the anion will make hydrolisis that can help us, to determine pH.

The  NH₄ClO₄ comes from a weak acid and a weak base. This is an special case.

a. KNO₂ →  K⁻  +  NO₂⁻

NO₂⁻  +  H₂O  ⇄  HNO₃  +  OH⁻        Kb = 1.41×10⁻¹¹

Kb = x² / (0.12 - x)

We can avoid the x in denominator because Kb it is a very low value. This expression is reduced to:

[OH⁻] = √(Kb . 0.12)

[OH⁻] = √(1.41×10⁻¹¹ . 0.12)  = 1.30×10⁻⁶ M

- log [OH⁻] = pOH → - log 1.30×10⁻⁶  = 5.88

14 - pOH = pH → 14 - 5.88 = 8.12

b. NaOCl  →  Na⁻  +  OCl⁻

OCl⁻  +  H₂O  ⇄  HClO  +  OH⁻        Kb = 3.33×10⁻⁷

The same as before, we avoid the x in denominator because Kb it is a very low value.

[OH⁻] = √(Kb . 0.37)

[OH⁻] = √(3.33×10⁻⁷ . 0.37)

[OH⁻] = 3.51×10⁻⁴ M

- log 3.51×10⁻⁴ = 3.45 → pOH

14 - 3.45 = 10.55 → pH

c. NH₄ClO₄ → NH₄⁺  +  ClO⁴⁻

Both ions can make hydrolisis in water. Ammonium can give ammonia and the hypochlorite makes hypochlorous acid.

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺      Ka = 5.7×10⁻¹⁰

OCl⁻  +  H₂O  ⇄  HClO  +  OH⁻        Kb = 3.33×10⁻⁷

The reduced expression to calculate pH for this sort of salts, is to make the square root of the product between the Ka constant.

As we have Kb for the HClO, we need to determine the Ka

Kw = Ka . Kb   →  Kw/Kb = Ka → 1×10⁻¹⁴ / 3.33×10⁻⁷ = 3×10⁻⁸

[H₃O⁺] = √(5.7×10⁻¹⁰ . 3×10⁻⁸) = 4.14×10⁻⁹ M

- log [H₃O⁺] = pH → - log 4.14×10⁻⁹ = 8.38

As per the calculations of the pH the solutions of 0.12 M KNO2 and 0.37 M NaOCl and the 0.39 M NH4ClO4 web assign have been made with a significant check on the figures.

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The strength of the intermolecular forces between diethyl ether molecules is weaker compared to the ion-dipole interactions between NaCl and diethyl ether.

In a solution of NaCl in diethyl ether \(\[ \text{{CH}}_3\text{{CH}}_2\text{{OCH}}_2\text{{CH}}_3 \]\)  the following intermolecular forces can be identified:

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Answer:

Of the following substances, PCl₃ has the highest boiling point

Explanation:

The given substances and their boiling point from online resources are;

Krypton, Kr, boiling point = -153.4°C

Chlorine gas, Cl₂, boiling point = -34.6°C

Borane, BH₃, (transient compound), boiling point of the dimer it forms B₂H₆ = -164.85°C

Methane, CH₄, boiling point = -161.6°C

Phosphorus trichloride, PCl₃, boiling point = 76.1°C

Therefore, given that PCl₃ is a volatile liquid at room temperature, while the other substances are gases and that PCl₃ has the highest boiling point of all the substances of 76.1°C, the substance with the highest boiling point is PCl₃.

The pH of the solution prepared by mixing 550.0 ml of 0.703 M CH₃COOH with 460.0 ml of 0.905 M NaCH₃COO is 4.745 (approx.).

To calculate the pH of the solution, we need to first find the concentration of acetic acid and acetate ion in the mixed solution. Then we can use the Henderson-Hasselbalch equation to determine the pH.

First, we find the moles of CH₃COOH and NaCH₃COO using the formula: moles = concentration x volume.

Moles of CH₃COOH = 0.703 M x 0.550 L = 0.38765 moles

Moles of NaCH₃COO = 0.905 M x 0.460 L = 0.4163 moles

Next, we calculate the concentrations of CH₃COOH and CH₃COO⁻ in the mixed solution.

[CH₃COOH] = (moles of CH₃COOH)/(total volume of solution) = 0.803 M

[CH₃COO⁻] = (moles of CH₃COO⁻)/(total volume of solution) = 0.683 M

Finally, we use the Henderson-Hasselbalch equation:

pH = pKa + log([CH₃COO⁻]/[CH₃COOH])

pKa = -log(Ka) = -log(1.76 × 10⁻⁵) = 4.753

pH = 4.753 + log(0.683/0.803) = 4.745

Therefore, the pH of the mixed solution is approximately 4.745.

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Answer:

Burning wood

Explanation:

Burning wood is a chemical change. Dissolving sugar is a physical change because it is still sugar, just dissolved. A broken bottle a physical change because it is just broken. Melting wax is a physical change because it is still wax, just melted. Burnt wood is a chemical change because it is now ash

Answer:

breaking a bottle

Explanation:

The physical mechanism causing the change in strength when using brine as the pore fluid is the presence of salt ions that weaken the interparticle bonds. Salt can reduce the shear strength of clays by increasing the repulsive forces between clay particles.

The strength of clay types at 50% water content can vary depending on whether DI water or brine is used as the pore fluid. Generally, there is a significant difference in strength between the two.

The presence of salt in brine can have an effect on the shear strength of clays. When salt is dissolved in water, it creates ions that can interact with the clay particles. These interactions can lead to the formation of electrical double layers around the clay particles, which can increase the interparticle repulsion and decrease the shear strength of the clay.

On the other hand, when DI water is used as the pore fluid, there is no presence of salt ions to affect the interparticle interactions. As a result, the clay particles can have stronger bonds and higher shear strength compared to when brine is present.

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Answer:

\(C_2H_4O_2\)

Explanation:

¡Hola!

En este caso, dado que conocemos la composición porcentual del compuesto, es posible primero obtener la formula empírica al asumir que dichos porcentajes son gramos, que se vuelven moles con las masas atómicas de carbono, hidrógeno y oxígeno respectivamente:

\(n_C=40gC*\frac{1molC}{12.01 gC}=3.33molC \\\\n_H=6.7gH*\frac{1molH}{1.01gC}=6.6molH\\\\n_O=53.3gO*\frac{1molO}{16.00gO}=3.33molO\)

De este modo, ahora obtenemos las relaciones molares entre ellos, con el fin de obtener los subíndices en la fórmula empírica:

\(C:\frac{3.33}{3.33}=1\\\\ H:\frac{6.6}{3.33}=2\\\\O:\frac{3.33}{3.33}=1\)

Por consiguiente, la fórmula empírica es:

\(CH_2O\)

Finalmente, dado que la masa molar del compuesto es 6.0/0.10=60g/mol (dada la masa y las moles), es posible notar que como la masa molar de la fórmula empírica es 30; esta es dos veces la molecular, por lo que esta ultima resulta:

\(C_2H_4O_2\)

¡Saludos!

The electron configuration of aluminum is 1s2 2s2 2p6 3s2 3p1.

One enzyme complex in the electron transport chain that utilizes iron-sulfur centers to transport electrons is the Complex I, also known as NADH dehydrogenase or NADH: ubiquinone oxidoreductase.

Complex I is a large multi-subunit enzyme complex located in the inner mitochondrial membrane. It plays a crucial role in the electron transport chain, which is responsible for generating ATP, the energy currency of the cell. Complex I accepts electrons from NADH, a molecule produced during the breakdown of glucose and other nutrients, and transfers them to ubiquinone (Coenzyme Q).

Within Complex I, there are multiple iron-sulfur clusters, which are prosthetic groups composed of iron and sulfur atoms. These clusters act as electron carriers, shuttling electrons along a series of redox reactions. The iron-sulfur centers within Complex I undergo reversible oxidation and reduction, allowing the transfer of electrons from NADH to ubiquinone. This process is crucial for establishing an electrochemical gradient across the mitochondrial membrane, which drives the synthesis of ATP.

In summary, Complex I is an enzyme complex in the electron transport chain that utilizes iron-sulfur centers to shuttle electrons. These iron-sulfur clusters play a vital role in the transfer of electrons from NADH to ubiquinone, contributing to the production of ATP within the cell.

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All the choices in the question are sources of error i.e.

TYPES OF ERROR:

Error can either be systematic or random.

According to this question, the following can be sources of errors in an experiment:

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Answer:

yes

Explanation:

because the atmosphere is the shield of earth

Answer:

they are seed producing so can populate more areas

Answer:

The correct answer is B. 1-propanamine.

Explanation:

Amines can form hydrogen bonds with water molecules, which makes them soluble in water. The more hydrogen bonding sites an amine has, the more soluble it will be in water.

Out of the given options, 1-propanamine has only one carbon chain, which allows it to form more hydrogen bonds with water molecules compared to amine molecules with longer carbon chains. Also, it does not have any other functional groups that could interfere with hydrogen bonding. Therefore, 1-propanamine would be the most soluble in water.

Hope it helps!

That solution of a compound which when dissolved in water conducts electricity, turns litmus red, and has a sour taste is C3H5(COOH)3 ( carboxylic acid ).

Carboxylic acid is can conduct electricity, turns blue litmus paper to red color and also has sour taste simply because it is a an acid; specifically, a weak acid. This is because when it dissolves in water, some of it hydrogen ions ionizes partially to produce positively charged hydrogen ions.

However, acid can be defined as a substance which when dissolved in water produces hydrogen ions as the only positive ions. All acids have sour taste, are corrosive when highly concentrated, turns blue litmus paper red and so on and so forth.

In conclusion, we can therefore conclude that carboxylic acid is a weak acid and also has characteristic properties of acids.

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Cells fit long pieces of DNA into the nucleus through a process called DNA packaging or DNA condensation.

In order to fit DNA into the small compartment of the nucleus, cells use specialized proteins called histones. DNA wraps around these histone proteins, forming structures known as nucleosomes. These nucleosomes further coil into a helical fiber called chromatin, which condenses even more to form chromosomes during cell division.

The process of DNA packaging allows cells to efficiently store and manage vast amounts of genetic information within the limited space of the nucleus. This compact arrangement also enables precise regulation of gene expression, as certain parts of the DNA can be made accessible or inaccessible for transcription based on cellular needs.

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Answer:

KOH + H2SO4 --> K2SO4 + H2O

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Effusion is the process by which a gas escapes through a tiny hole or porous material into a region of lower pressure. The rate of effusion is directly proportional to the average speed of the gas particles, which in turn is related to their molecular weight. The ratio of the rate of effusion of CO₂ to that of He is 0.302.

The rate of effusion is inversely proportional to the square root of the molar mass of a gas. The molar mass of CO₂ is 44.01 g/mol, while the molar mass of He is 4.00 g/mol. Thus, the square root of the molar mass of CO₂ is √44.01 = 6.63 g/mol, and the square root of the molar mass of He is √4.00 = 2.00 g/mol.

Using the equation for the ratio of the rates of effusion, we get:

Rate of CO₂ effusion / Rate of He effusion = √(Molar mass of He) / √(Molar mass of CO₂)

= 2.00 / 6.63 = 0.302

Therefore, the ratio of the rate of effusion of CO₂ to that of He is 0.302.

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Answer:

increase the rate of reaction

If we add 25.6 g of an unknown molecular compound to the 0.250 kg of benzene, the molar mass of the unknown compound is 148.8 g/mol.

The Molality of the compound is given as :

ΔT = i Kf m

Where,

ΔT = freezing point depression = 3.54 °C

i =  Van't Hoff factor of the Benzene = 1

Kf =  constant of the freezing = 5.12 °C/m

m = molality = ?

m = ΔT / i Kf

m = 3.54 / 1 × 5.12

m = 0.69 mol

molality = moles / mass of benzene

moles = 0.172

The molar mass = mass / moles

The molar mass = 25.6 / 0.172

The molar mass = 148.8 g/mol

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Answer:

To produce a solution that frees at -50°C we need 1668.5 g of ethylene glycol

Explanation:

Formula for freezing point depression is:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m . i

Our solute is the ethylene glycol, then water is our solvent.

As ethylene glycol is a non ionic compound, i = 1 (Van't Hoff factor, numbers of ions dissolved)

Kf = Cryoscopic constant. For water is 1.86 °C/m

We replace data:

0°C - (-50°C) = 1.86 °C/m . m . 1

To determine grams of ethylene glycol needed, we need to find m (molality)

50°C / 1.86 m/°C = m → 26.8 mol/kg

As this moles of solute are contained in 1kg, definetely we need 26.8 moles of ethylene glycol.

To find the answer, we convert moles to mass:

26.8 mol . 62.07g / mol = 1668.5 g

Answer:

56.9 amu

Explanation:

Depending on the number of isotopes an element has, the Average atomic mass of isotopes of an element can be calculated using:

(atomic mass of isotope1 × decimal abundance of isotope1) + (atomic mass of isotope2 × decimal abundance of isotope2)

According to this question, two isotopes of element Z were given

- Isotope Z- 56 with atomic mass 56 is 70% abundance. This means that its decimal abundance is 0.70

- Isotope Z- 59 with atomic mass 59 is 30% abundance. This means that its decimal abundance is 0.30

Therefore, average atomic mass of element Z is:

= (56 × 0.70) + (59 × 0.30)

= (39.2) + (17.7)

= 56.9

Approximately 0.323 half-lives have occurred over the period of 1.00 hour.

To determine the concentration of N₂O₅ after 1.00 hour and the number of half-lives that occurred, we can use the first-order rate equation and the concept of half-life.

The first-order rate equation is given by:

ln([A]t/[A]₀) = -kt

Where [A]t is the concentration at time t, [A]₀ is the initial concentration, k is the rate constant, and t is the time.

Given that k = 6.22 x 10^(-5) s^(-1), [A]₀ = 0.500 M, and t = 1.00 hour = 3600 seconds, we can substitute these values into the equation.

ln([A]t/0.500) = -(6.22 x 10^(-5))(3600)

ln([A]t/0.500) = -0.22392

[A]t/0.500 = e^(-0.22392)

[A]t = 0.500 * e^(-0.22392)

[A]t ≈ 0.500 * 0.7997

[A]t ≈ 0.3999 M

Therefore, the concentration of N₂O₅ after 1.00 hour is approximately 0.3999 M.

To calculate the number of half-lives, we can use the equation:

t₁/2 = ln(2)/k

t₁/2 = ln(2)/(6.22 x 10^(-5))

t₁/2 ≈ 11126.03 seconds

Number of half-lives = t/t₁/2

Number of half-lives = 3600/11126.03

Number of half-lives ≈ 0.323

Therefore, approximately 0.323 half-lives have occurred over the period of 1.00 hour.

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Dinitrogen pentoxide decomposes to form nitrogen dioxide and oxygen gas according to the following equation: 2 N₂0₅ → 4 NO₂+ O₂ Experimentally, this process is found to obey first order kinetics, with k 6.22 x 10 s 1. f the initial concentration of N205 is 0.500 M, what will its concentration be after 1.00 hour? How many half lives have occurred over this period of time?

The balanced equation for the given chemical reaction is:

Cu(s) + 4 HNO3(aq) → Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)

In this chemical reaction, copper (Cu) reacts with nitric acid (HNO3) to produce copper nitrate, nitrogen dioxide, and water.

A balanced equation is one where the number of atoms on both sides of the equation is the same. The balanced equation shows that one copper (Cu) atom reacts with four nitric acid (HNO3) molecules, forming one copper nitrate (Cu(NO3)2) molecule, two nitrogen dioxide (NO2) molecules, and two water (H2O) molecules.

Thus, when the given equation is balanced with the lowest whole-number coefficients, the coefficients are:

Cu(s) + 4 HNO3(aq) = Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l).

Therefore, the coefficients of the equation are Cu(s) + 4HNO3(aq) + Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l).

The solution to the problem above was provided with the correct balanced equation and a thorough explanation.

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Answer:

The final temperature of water is 34.92 ⁰C

The specific heat of the metal is 2357.78 J/kg⁰C

Explanation:

Given;

mass of the hot metal, m = 10.0 g = 0.01 kg

temperature of the hot metal, \(t_m\) = 300 °C

mass of water, \(m_w\) = 150 g = 0.15 kg

initial temperature of the water, \(t_w_i\) = 25.0 °C

heat lost by the hot metal = heat gained by water

Q = 6.25 kJ = 6250 J

let the final temperature of water = T

\(Q = m_w C_p_w (T-t_w_i)\\\\T-t_w_i = \frac{Q}{m_wC_p_w}\\\\ T= \frac{Q}{m_wC_p_w} + t_w_i\\\\T = \frac{6250}{(0.15)(4200)} + 25\\\\T = 34.92 ^0 C\)

The final temperature of water is also the equilibrium temperature.

The specific heat of the metal is given by

\(Q = mC_p_m (300 - T)\\\\C_p_m = \frac{Q}{m(300 - T)}\\\\ C_p_m = \frac{6250}{0.01(300 - 34.92)}\\\\ C_p_m = 2357.78 \ J/kg ^0C\)

Answer: When you add an electron to zinc, it needs some extra energy. This is because zinc atoms naturally don't like having an extra electron. The extra electron and the electrons already present in zinc repel each other due to their negative charges. So, you have to give some energy to the zinc atom to overcome this repulsion and make it accept the additional electron. Basically, energy input is required to make zinc accept an extra electron because the electron doesn't fit easily and needs some force to be added.

Explanation: hope this helps

Answer:

The pH scale measures acidity of a substance. known as potential of hydrogen, it varies from 0 to 14 with 7 being the pH value of a neutral solution. Below 7 shows the substance is acidic in nature and above 7 is alkaline in nature. pH 0-3 are considered strong acids while pH 4-6 are weak acids. pH 8-10 are weak alkalines and pH 11-14 are strong alkalines. This is a general trend and there may be exeptions especially if the substance has a negative pH. However, it would not be covered likely unless you are doing university chemistry.