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If a substance has a high PH, it will feel__________.

If a substance has low PH, it will taste__________.

The volume of 6.9 mol of oxygen at 233 K and a pressure of 4.0 atm is approximately 12.0L.

To calculate volume of a gas, we can make use of Ideal Gas Law equation. It is a fundamental equation in thermodynamics that describes the behaviour of an ideal gas under certain circumstances. It relates pressure(P), volume(V), number of moles (n), and temperature(T) of an ideal gas using the equation:

PV = nRT

Where P = Pressure of the gas,

           V = Volume of the gas,

           n = Number of moles of the gas,

           R = Ideal gas constant commonly expressed as 0.0821  L·atm/(mol·K) or 8.314 J/(mol·K),

          T = Temperature of the gas.

In the question, we are given with:

n = 6.9 mol

T = 233 k

P = 4.0 atm

Substituting the above values in the equation to find the volume, we get:

4.0 * V = 6.9 * 0.0821 * 233

V = (6.9 * 0.0821 * 233) / 4.0

V = 11.9997 (approximately 12.0)

Therefore, The volume of 6.9 mol of oxygen at 233 K and a pressure of 4.0 atm is approximately 12.0L.

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If the patient is not undergoing dialysis, the conclusion drawn from the results given is that one of the values is in error.

The patient's creatine is approximately five times the upper normal range (0.6-1.2 mg/dL), whereas the patient's urea nitrogen is within the normal range (6-20 mg/dL). When there is kidney damage, significant increases in urea nitrogen nearly usually go hand in hand with significant increases in creatinine.

The ratio of urea to creatinine, which is typically 10:1 to 20:1, is extremely low in our case at 3:1. As a result, either the creatinine or urea nitrogen measurement in this instance is wrong. Repeating both tests is necessary.

Dialysis is a method used to purify or filter your blood through the used of a needle and a dialyzer. It is used when your kidney does not work properly or are damaged.

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Gamma, alpha, and beta radiation are all forms of ionizing radiation emitted during radioactive decay, but they differ in terms of their components, energy levels, examples, creation, and safety in various types of nuclear energy.

Gamma radiation consists of high-energy photons, similar to X-rays. It possesses the highest energy level among the three types and can penetrate several centimeters of lead or several meters of concrete.

Examples of gamma-emitting isotopes include cobalt-60 and cesium-137. Gamma rays are created during nuclear reactions and decay processes, such as fission or fusion reactions. They pose a significant risk to human health due to their ability to damage living tissue, but their penetration power makes them useful in medical imaging and cancer treatment.

Alpha radiation consists of alpha particles, which are composed of two protons and two neutrons (helium nuclei). They have low energy levels and can be stopped by a sheet of paper or a few centimeters of air.

Examples of alpha-emitting isotopes include uranium-238 and radon-222. Alpha particles are created through the decay of heavy elements. While they can cause significant damage if inhaled or ingested, they are less penetrating and therefore less hazardous outside the body.

Beta radiation involves the emission of beta particles, which are high-energy electrons (beta-minus) or positrons (beta-plus). They have moderate energy levels and can penetrate several millimeters of aluminum.

Examples of beta-emitting isotopes include carbon-14 and strontium-90. Beta particles are created during the decay of certain isotopes, where a neutron is transformed into a proton or vice versa. Beta radiation poses an intermediate level of risk, as it can penetrate the skin and cause tissue damage, but it is less harmful than gamma radiation.

In terms of nuclear energy, gamma radiation is a concern in all types of reactors, as it is released during fission and fusion reactions. Shielding is necessary to protect workers and the environment.

Alpha radiation is of particular concern in nuclear fuel cycle processes like uranium mining and enrichment. Beta radiation is relevant in nuclear power plant operations, as some fission products emit beta particles. It requires appropriate shielding and monitoring to ensure worker safety.

Overall, gamma radiation has the highest energy, alpha radiation has the lowest, and beta radiation falls in between. Their differing penetration abilities, creation mechanisms, and safety considerations make them suitable for various applications and require tailored safety measures.

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The gas given off at the cathode is hydrogen while the gas given off at the anode is oxygen.

The term electrolysis has to do with the decomposition of a substance by the passage of direct current through it.

The ions present in the system are H^+, OH^- and NO3^- ions. The gas given off at the cathode is hydrogen while the gas given off at the anode is oxygen.

The half reaction equation of the reduction is;

2H^+(aq) + e ----->H2(g)

The oxidation half reaction equation is;

4OH^- (aq)------> 2H2O(l) + 4e

The overall equation is;

4H^+(aq) + 4OH^- (aq)------> 4H2(g) + 2H2O(l) + 4e

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Answer: try 42.8

Explanation: you might have to round to 42.8 because of the significant  figures

Answer:

it reacts to limestone better and is a bit more firmer and mostly preferred over because of how fast it can react towards chemicals

The subscript 4 in [Cu(NH3)4]2+ indicates the B)coordination number of Cu2+.

Coordination number refers to the number of ligands attached to the central metal ion in a coordination complex. In the given complex, Cu2+ is the central metal ion and it is coordinated to four ammonia (NH3) ligands.

The subscript 4 in [Cu(NH3)4]2+ indicates the number of NH3 ligands attached to the Cu2+ ion, which is the coordination number of Cu2+. The oxidation number of Cu in this complex is +2, which is indicated by the Roman numeral II in the formula. Therefore, the correct answer is B.

Answer:

Explanation:

Balanced Chemical Equation:

                            Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag

Step 1: Find Limiting Reagent

Moles of Cu:

                   Mole = 3 g / 63.55 g/mol = 0.04721 mol

Moles of AgNO₃:

                   Moles = 3.85 g / 169.87 g/mol = 0.022664 mol

Also, Mole ratio of Cu : AgNO₃ is 1 : 2. Hence, 0.04721 mol of Cu will require 0.09442 mol of AgNO₃. This means, AgNO₃ is the limiting reagent.

Step 2: Calculate moles of Ag:

0.022664 mol of AgNO₃ will produce 0.022664 mol of Ag because both have same ratio in balanced chemical equation.

Step 3: Calculate mass of Ag:

                      Mass = 0.022664 mol × 107.87 g/mol = 2.4447 g

The work done by the gas is -8.00 L.atm.

Given data:

Initial volume = 8.00 L

Final volume = 16.0 L

External pressure = 1.00 atm

The equation for the work done by a gas during a reversible, isothermal process is given by: W = - nRT \ln \frac{V_f}{V_i}

where n is the number of moles of the gas, R is the gas constant (8.31 J/(mol.K) for ideal gases), T is the temperature in Kelvin, V_i and V_f are the initial and final volumes of the gas respectively.

Since the pressure of the gas is constant, the process is isobaric (constant pressure).

Therefore, W = - P \Delta V$$where P is the external pressure and $\Delta V = V_f - V_i is the change in volume. ,

Substituting the given values into the above equation, we haveW = - (1.00 \; \text{atm}) \times (16.0 \; \text{L} - 8.00 \; \text{L})W = -8.00 \; \text{L.atm}

Hence, the work done by the gas is -8.00 L.atm.

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The major organic product obtained from the addition reaction of HCl to 3-methyl-1-butene is 2-chloro-3-methylbutane. Option d is correct.

The reaction mechanism involves the electrophilic addition of HCl to the double bond of the 3-methyl-1-butene, leading to the formation of a carbocation intermediate. The H⁺ ion from HCl acts as an electrophile, attacking the double bond and forming a carbocation on the tertiary carbon. The chloride ion then acts as a nucleophile, attacking the carbocation and forming the final product.

The addition of HCl to the double bond results in the formation of two possible carbocation intermediates, one on the primary carbon and the other on the tertiary carbon. However, the more stable tertiary carbocation intermediate is formed faster and is favored under the reaction conditions. The chloride ion then attacks the tertiary carbocation, resulting in the formation of the product, 2-chloro-3-methylbutane. Option d is correct.

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Answer:

The empirical formula simply states the ratios of the elements that make up the molecule, while molecular formulas specify the amounts of each element exactly, not just giving the ratio. Let's take ethane (C2H6) as an example.

Answer:

0.4193924945 million

Explanation:

3.5(0.119826427)

Answer:

13.33 g/dm³

Explanation:

Concentration (g/dm³)= mass(g) ÷ volume (dm³)

Now you need to convert 150 cm³ to dm³

1000cm³ = 1 dm³

thus, 150 cm3= 150 ÷ 1000

= 15dm³

and you already have mass in grams

so concentration  = 2 ÷ 0.15

= 13.33 g/dm³ and there you go.. solved ;)

The rate of effusion of CO₂ under the same conditions is approximately 6.21 mole/min. Option B.

According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, we can use the ratio of molar masses to determine the rate of effusion of CO₂ (carbon dioxide) compared to NH₃ (ammonia).

The molar mass of NH₃ is approximately 17.03 g/mol, while the molar mass of CO₂ is approximately 44.01 g/mol.

Let's denote the rate of effusion of CO₂ as x mole/min. Using the ratio of molar masses, we can set up the following proportion:

(√molar mass of NH₃) / (√molar mass of CO₂) = rate of effusion of NH₃ / rate of effusion of CO₂

√17.03 / √44.01 = 2.40 mole/min / x mole/min

Solving for x, we find:

x = 2.40 mole/min * (√molar mass of CO₂) / (√molar mass of NH₃)

= 2.40 mole/min * (√44.01 g/mol) / (√17.03 g/mol)

≈ 6.21 mole/min

Therefore, the rate of effusion of CO₂ under the same conditions is approximately 6.21 mole/min.

The correct answer is (b) 6.21.

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Answer:

Nitrogen is an inert gas — meaning it doesn't chemically react with other gases — and it isn't toxic. But breathing pure nitrogen is deadly. That's because the gas displaces oxygen in the lungs. Unconsciousness can occur within one or two breaths, according to the U.S. Chemical Safety and Hazard Investigation Board.

Answer:

because it is an insert gas and it is very dangerous for ear and our nose

Answer:

Explanation:

waht is the solo0urtsin

the answer is Te because O is oxygen.

Answer:

O

Explanation:

this is because it is oxygen

Answer: I don’t know lol

Explanation:  I am so sorry I thought this was easy

Answer:

Ant.

Explanation:

The dog would be the most frantic, when all are compared at an equal speed of 30 mph. An ant of such strength, has six jointed legs and has an exoskeleton. An Ant of such size(mass) and weight would have take off with better traction too(6 legs). So the Ant would have the greatest kinetic energy. Also it would take turns while fleeing from the wildfire, while running at almost 30 mph while the other animals wouldn't come that close.

It is technically and literally illogical to say an ant can do 30 mph in its regular size.

Dont believe me? Watch ants scurrying away when threatened and taking turns and U-turns. Six legs offer more energy while fleeing by instinct. Literally six legs at 30 mph. Crazy!

To calculate the pH of the solution, we need to consider the dissociation of the trimethylamine (base) and the trimethylammonium chloride (conjugate acid) in water. From this, the pH of the solution containing comes out to be approximately 9.73.

The dissociation reaction of trimethylamine can be represented as follows:

(CH₃)₃N + H₂O ⇌ (CH₃)₃NH⁺ + OH⁻

The dissociation reaction of trimethylammonium chloride can be represented as follows:

(CH₃)₃NH⁺Cl⁻ + H₂O ⇌ (CH₃)₃NH⁺ + Cl⁻ + H₂O

Since the concentration of the trimethylammonium chloride is higher than the concentration of trimethylamine, we can assume that all the trimethylamine has been protonated to form (CH₃)₃NH⁺. Therefore, we can consider the concentration of (CH₃)₃NH⁺ to be equal to the concentration of trimethylammonium chloride, which is 0.13 M.

Now, we need to calculate the concentration of hydroxide ions (OH⁻) in the solution. The concentration of hydroxide ions can be determined using the equilibrium constant for the reaction of trimethylamine with water, which is the Kb value.

The Kb value for trimethylamine is usually given as 6.3 x 10⁻⁵ at a certain temperature. However, since the Kb value was not specified in the question, we will assume a generic value of Kb = 1.0 x 10⁻⁴ for illustrative purposes.

Using the Kb value and the concentration of trimethylamine (0.070 M), we can calculate the concentration of hydroxide ions (OH⁻) using the equilibrium expression for the base dissociation:

Kb = [OH⁻][ (CH₃)₃NH⁺] / [(CH₃)₃N]

Since all the trimethylamine has been protonated, the concentration of (CH₃)₃N is negligible compared to the concentration of (CH₃)₃NH⁺. Therefore, we can approximate the equilibrium expression as:

Kb = [OH⁻][ (CH₃)₃NH⁺] / 0.070 M

Now, we can solve for [OH⁻] using the Kb value and the concentration of trimethylamine:

1.0 x 10⁻⁴ = [OH⁻] x 0.13 M / 0.070 M

Simplifying the equation:

[OH⁻] = (1.0 x 10⁻⁴) x (0.070 M) / (0.13 M)

[OH⁻] ≈ 5.38 x 10⁻⁵ M

Now, we can calculate the pOH of the solution using the concentration of hydroxide ions:

pOH = -log10([OH⁻])

pOH ≈ -log10(5.38 x 10⁻⁵)

pOH ≈ 4.27

Finally, we can calculate the pH of the solution using the relation:

pH = 14 - pOH

pH ≈ 14 - 4.27

pH ≈ 9.73

Therefore, the pH of the solution containing 0.070 M trimethylamine and 0.13 M trimethylammonium chloride is approximately 9.73.

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To achieve a temperature close to -15 °C for cooling the reaction mixture, the best cooling method would be an acetone/dry ice bath or a liquid nitrogen bath.

The temperature required for cooling the reaction mixture to -15 °C is quite low and may not be easily achieved using common methods like an ice water bath or an ice/NaCl bath. These methods typically reach temperatures around 0 °C or slightly below freezing.

An acetone/dry ice bath is a suitable cooling method as the combination of acetone and dry ice can reach temperatures as low as -78 °C. This temperature range is sufficient to cool the reaction mixture to -15 °C effectively.

Alternatively, a liquid nitrogen bath is another excellent option. Liquid nitrogen has an extremely low boiling point of -196 °C, making it capable of achieving and maintaining temperatures well below -15 °C.

Setting the reaction in a freezer may not be the best option as the temperature control in a standard freezer may not reach the desired -15 °C, and there could be limited space or other factors that might affect the reaction conditions. Therefore, an acetone/dry ice bath or a liquid nitrogen bath would be the preferred cooling methods for achieving a temperature close to -15 °C.

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Answer:

There are 0.5 mole in 20g of argon.

Explanation:

40 g of argon = 1mole

Then 20g of argon is,

→ 1/40 × 20

→ 0.5 mole

Answer:

\(n = \frac{mass}{molar \: mass} \\ n = \frac{20g}{39.948 \frac{g}{mol} } \\ n = 0.5 \: mol \: of \: argon\)

Explanation:

hope this makes sense

:)

To determine the amount of copper plated out by a current of 2.3 A, we need to consider Faraday's law of electrolysis. By applying the formula Q = It, where Q is the quantity of electricity, I is the current, and t is the time.

Faraday's law of electrolysis states that the mass of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The formula Q = It represents the relationship, where Q is the quantity of electricity in coulombs, I is the current in amperes, and t is the time in seconds.

To calculate the mass of copper plated out, we need to know the molar mass of copper, which is approximately 63.55 g/mol. By rearranging the formula Q = It to solve for Q, we have Q = I × t. Substituting the given current (2.3 A) and the time, we obtain the quantity of electricity in coulombs.

Finally, to find the mass of copper, we divide the quantity of electricity by the Faraday constant, which is 96,485 coulombs per mole. Multiplying this result by the molar mass of copper gives us the mass of copper plated out.

Therefore, by applying Faraday's law of electrolysis, a current of 2.3 A will plate out a certain mass of copper, which can be calculated using the formula Q = It and considering the molar mass of copper.

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Answer:

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

\(n = \frac{N}{L} \\\)

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

We have

\(n = \frac{2.4 {10}^{23} }{6.02 \times {10}^{23} } = \frac{2.4}{6.02} \\ = 0.3986...\)

We have the final answer as

Hope this helps you

The overall formation constant is 1.11*10¹³.

What is ion ?

An atom or collection of atoms that is classified as an ion has a different number of electrons than proton atoms.

What is solubility ?

A substance's solubility is the greatest amount that will dissolve in a given amount of solvent at a given temperature. A specific solute-solvent combination's solubility is a defining trait, and the solubilities of various compounds can vary significantly.

The complex [Ni(en)3]2+ is more stable than the complexes [Ni(NH3)6]2+ and [Ni(H2O)6]2+.

Steric hindrance of ligands will generally reduce the value of the formation constant.

The coordination number of the central metal ion is related to the formation constant of the complex.

Therefore, the overall formation constant is 1.11*10¹³.

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Answer:

The third option should be right

Answer:

C.

Explanation:

edg. 2020 just took the test

After dissolving in solution, the following a strong acid dissociates completely in solution to produce its conjugate and the conjugate of the strong acid is the ion this applies. Here options B and D are the correct answer.

A strong acid is one that completely dissociates into its constituent ions in an aqueous solution. This means that all of the acid molecules break apart into hydrogen ions (H+) and the corresponding anions. Therefore, option B is correct, while option A is incorrect.

The conjugate base of a strong acid is an anion because the hydrogen ion (H+) has been removed from the acid molecule. This anion may be neutral or basic in solution, depending on the identity of the anion. Therefore, option E is correct, while options C and D are incorrect. Finally, the conjugate of a strong acid is not a cation, so option F is also incorrect.

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Complete question:

Which of the following applies to a strong acid once it dissolves in solution?

A. It dissociates partially in solution to produce its conjugate

B. It dissociates completely in solution to produce its conjugate

C. The conjugate of a strong acid is neutral in pH when in solution

D. The conjugate of a strong acid is basic in the solution

E. The conjugate of a strong acid is an anion

F. The conjugate of a strong acid is a cation

Answer:

1. Unsaturated

2. Alkanes

3. alkynes

4. gasoline and lubricating oils.

5. manufacturing plastic products

The equilibrium concentration of OAc-, if  500 mL of a 0.100 M solution of HOAc(aq) is prepared, is approximately 1.33 x 1\(0^{-3}\) M.

To determine the equilibrium concentration of OAc- in the solution, we can use the acid dissociation constant (Ka) and an ICE (Initial, Change, Equilibrium) table. The reaction for the dissociation of HOAc is:

HOAc(aq) ⇌ \(H^{+}\)(aq) + OAc-(aq)

Initially, the concentrations are [HOAc] = 0.100 M, [\(H^{+}\)] = 0, and [OAc-] = 0. Let x be the change in concentration for dissociation. At equilibrium, we have:

[HOAc] = 0.100 - x
[\(H^{+}\)] = x
[OAc-] = x

Now, using the given Ka value (1.79 x 1\(0^{-5}\)):

Ka = ([\(H^{+}\)][OAc-])/[HOAc] = (x * x) / (0.100 - x)

Solving the quadratic equation, x ≈ 1.33 x 1\(0^{-3}\) M, which represents the equilibrium concentration of both H+ and OAc-. Therefore, the equilibrium concentration of OAc- is approximately 1.33 x 1\(0^{-3}\) M.

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Answer:

Explanation:

The only answer you can get would be that the area was once covered with water.

A glacier wouldn't allow the fish to move

Predators would likely eat more than they would leave behind as evidence.

If they developed limbs, at some of them would show up that way.