At a fixed volume, a four-fold increase in the temperature of a gas will lead to _______ in pressure.

A cation will not replace an anion in a compound because the two positively-charged ions will repel each other.

Ions are atoms or group of atoms possessing an electrical charge.

Ionic compounds are compounds which are composed of oppositely charged ions held together by strong electrostatic forces of attraction between the oppositely charged ions.

If a cation replaces an anion, the compound would have two l positive charges which will repel each other.

Therefore, a cation will not replace an anion in a compound because the two positively-charged ions will repel each other.

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Answer:

Cardiac muscle cells are located in the walls of the heart, appear striated, and are under involuntary control. Smooth muscle fibers are located in walls of hollow visceral organs, except the heart, appear spindle-shaped, and are also under involuntary control.

Explanation:

Answer:

My hypothesis is that: If the electrolyte source is changed (potato, apple, lime, lemon), then the production of energy (measured in volts) using a lemon will produce the highest voltage because the acid content in the fruit or vegetable will produce electricity when in contact with the electrodes (both zinc and copper

Explanation:

Answer:

Freezing point depression=2×1.027⋅mol⋅kg−1×1.86⋅K⋅kg⋅mol−1=3.82⋅K .. And so the fusion point of the solution will be −3.8 ∘C .

Approximately 70.57 mL of 1.00 M NaOH should be added to 250 mL of 0.50 M CH3CO₂H to produce a buffer with a pH of 4.50.

To determine the volume of 1.00 M NaOH required to produce a buffer with a pH of 4.50 when added to 250 mL of 0.50 M CH3CO₂H, we need to consider the Henderson-Hasselbalch equation and the stoichiometry of the reaction.

The Henderson-Hasselbalch equation for a buffer solution is given as:

pH = pKa + log([A-]/[HA])

In this case, CH3CO₂H (acetic acid) acts as the weak acid (HA) and CH3COO- (acetate ion) acts as its conjugate base (A-). We are given that the desired pH is 4.50, and we can determine the pKa value for acetic acid from reference sources, which is approximately 4.75.

Using the Henderson-Hasselbalch equation, we can rearrange it to solve for the ratio [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(4.50 - 4.75) = 10^(-0.25) = 0.5623

This means that the ratio of the acetate ion to acetic acid in the buffer solution should be approximately 0.5623.

To calculate the required volume of NaOH, we need to consider the stoichiometry of the reaction. Acetic acid reacts with hydroxide ions (OH-) to form acetate ions and water:

CH3CO₂H + OH- → CH3COO- + H2O

The stoichiometric ratio between acetic acid and hydroxide ions is 1:1. Therefore, the volume of 1.00 M NaOH needed can be calculated using the equation:

Volume (NaOH) × 1.00 M = Volume (CH3CO₂H) × 0.50 M × 0.5623

Volume (NaOH) = (Volume (CH3CO₂H) × 0.50 M × 0.5623) / 1.00 M

Volume (NaOH) = (250 mL × 0.50 M × 0.5623) / 1.00 M

Volume (NaOH) ≈ 70.57 mL

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Answer:

I would answer but not enough information

The experimental setup includes the variables that determine the result. The neighbor with a traditional garage door opener is the control, volunteers are independent, and the number of door openings is the dependent variable.

Dependent variables are said to be the data that are affected and altered by the other variable of the experimental design. In the experiment of the door opener app, the number of times the door opens is the dependent variable as it depends on the distance.

Independent variables are said to be the data that influence the dependent variable and are the cause in an experimental group are the test variables. The volunteers that have access to the door open app are the independent variables.

On the other hand, the control is the group that lacks the independent variable and is deemed to be constant as they must alter the other factor leading to changes in the result. The number of times the garage door opens is constant.

Therefore, the independent and dependent variables are the test variables.

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Answer: D.

Explanation: I got it right and the dude above me is spreading fake news

Answer:

\([A]=0.0026M\)

Explanation:

Hello.

In this case, since the second-order rate law is:

\(\frac{dC_A}{dt}=-kC_A^2\)

Whereas the subscript A accounts for C2F4 and its integration turns out into:

\(\frac{1}{[A]}= \frac{1}{[A_0]} +kt\)

Thus, for the initial concentration of C2F4 computed via the 0.1 mol in the 2.31-L container:

\([A]_0=\frac{0.1mol}{2.31L} =0.043M\)

The final concentration after 2.1 h is:

\(\frac{1}{[A]}= \frac{1}{0.043M} +\frac{0.048}{M*s} *\frac{3600s}{1h}*2.1h\\\\\frac{1}{[A]}=\frac{386.1}{M}\)

Solving for the final concentration of C2F4, we obtain:

\([A]=\frac{M}{386.1} =0.0026M\)

Best regards.

Answer:

2

Explanation:

You write your answer with the same number of significant figures as the number with the smallest amount of figures, in this case, that number is 2.1 so your answer should be written with 2 significant figures.

Answer:

Electronegativities generally increase from left to right across a period

Therefore the answer is B.

Explanation:

An orbital filling diagram is the more visual way to represent the arrangement of all the electrons in a particular atom. In an orbital filling diagram, the individual orbitals are shown as circles (or squares) and orbitals within a sublevel are drawn next to each other horizontally.~

The pressure of hydrogen gas is 276.25 Kpa .

Given,

Mass of aluminum = 29.51g

Temperature (T) = 25.67 degC =298.67 K

Volume (V) = 14.75 L

The required equation when aluminum reacts with excess sodium hydroxide and water is given by ,

2Al + 2NaOH + 2H2O ==>2NaAlO2 + 3H2

molecular mass of aluminum =26.98 g

1 mole of aluminum= 26.98g

2 moles of aluminum = 53.96g

2 mole of aluminum produces =  3  moles of hydrogen gas

53.96 g of aluminum produces = 6 g of hydrogen gas

29.51 g of aluminum produces = 6*(29.51) /53.96 =3.28 g of hydrogen gas

Thus ,

2 g of hydrogen gas = 1 mole of hydrogen

3.28 g of hydrogen gas = 3.28/2 mole =1.64 mol of hydrogen

Thus , n = 1.64 mol

According to ideal gas equation ,

PV=nRT

P=nRT/V

P = 1.64 * (0.0821L atm K^-1mol^-1 ) *(298.67 K)/14.75L

P=2.726 atm

P= 276.25 Kpa

Hence the pressure of hydrogen gas is 276.25 Kpa .

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The balanced chemical equation for the reaction between strontium and oxygen can be written as follows: 2 Sr (s) + \(O_2\)(g) → 2 SrO (s).

In this equation, solid strontium (Sr) reacts with gaseous oxygen (\(O_2\)) to produce solid strontium oxide (SrO).

To determine the mass of product expected to form in this reaction, we need to consider the molar ratio between strontium and strontium oxide. From the balanced equation, we can see that 2 moles of strontium react to produce 2 moles of strontium oxide.

The molar mass of strontium (Sr) is 87.62 g/mol, and the molar mass of strontium oxide (SrO) is 119.62 g/mol. Since the molar ratio is 1:1 between strontium and strontium oxide, the mass of strontium oxide formed will be equal to the mass of strontium used.

In this case, the student added 4.93 g of strontium to the crucible. Therefore, the expected mass of strontium oxide formed will also be 4.93 g.

It's important to note that this calculation assumes that the reaction proceeds to completion, meaning that all of the strontium reacts with oxygen. In actual laboratory conditions, the yield of the reaction may be less than 100% due to factors such as incomplete reaction, side reactions, or product loss.

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Answer:

A ---->gamma ray

Explanation:

Gamma rays have the highest frequencies among all electromagnetic waves and therefore have the shortest wavelengths.

Answer:

Wind energy

Explanation:

Answer: Wind energy

Explanation:

Approximately 631.5 liters of dry air at 25°C and 750 torr would need to be processed to produce 150 liters of liquid \(O_2\) -183°C.

To solve this problem, we need to consider the ideal gas law and the molar volume of gases.

First, we can calculate the number of moles of oxygen in 150 L of liquid \(O_2\) at -183°C. To do this, we divide the mass of liquid oxygen by its molar mass:

Mass of liquid oxygen = volume of liquid oxygen * density of liquid oxygen = 150 L * 1.14 g/mL = 171 g

Molar mass of oxygen (O2) = 32 g/mol

Number of moles of oxygen = mass of oxygen / molar mass of oxygen = 171 g / 32 g/mol ≈ 5.34 mol

Since the mole fraction of oxygen in dry air is given as 0.21, we can calculate the total moles of dry air needed to produce 5.34 mol of oxygen:

Moles of dry air = moles of oxygen / mole fraction of oxygen = 5.34 mol / 0.21 ≈ 25.43 mol

Now, we can use the ideal gas law to calculate the volume of dry air at 25°C and 750 torr (convert to atm) that corresponds to 25.43 mol:

PV = nRT

P = 750 torr * (1 atm / 760 torr) ≈ 0.987 atm

V = volume of dry air (unknown)

n = 25.43 mol

R = 0.0821 L·atm/(mol·K)

T = 25°C + 273.15 = 298.15 K

Solving for V:

V = nRT / P = (25.43 mol)(0.0821 L·atm/(mol·K))(298.15 K) / 0.987 atm ≈ 631.5 L

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Molarity = (moles of solute / volume of solution). Since the moles of solute (trypticase) stays the same, the molarity increases when the volume of solution decreases.

The volume of a word is an expression of the amount of content in a written piece. It can be measured in terms of the number of words, pages, or characters contained within a text. A text with a high volume will contain more content than a text with a low volume. Plagiarism-free writing is essential for a text to have a high volume, as plagiarized content does not contribute to the overall value of the piece.

The new molarity of trypticase in the cell growth medium would be 48 μM.

Molarity = (moles of solute / volume of solution)

Molarity (original) = (12 μM / 58 mL)

Molarity (new) = moles of solute (original) / volume of solution (new)

Molarity (new) = (12 μM * 58 mL) / 16 mL

Molarity (new) = 48 μM

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Explanation:

These are the first 20 elements, listed in order:

H - Hydrogen

He - Helium

Li - Lithium

Be - Beryllium

B - Boron

C - Carbon

N - Nitrogen

O - Oxygen

F - Fluorine

Ne - Neon

Na - Sodium

Mg - Magnesium

Al - Aluminum

Si - Silicon

P - Phosphorus

S - Sulfur

Cl - Chlorine

Ar - Argon

K - Potassium

Ca - Calcium

I HOPE IT HELPS YOU

Answer:

one from each parent

Explanation:

Answer:

your answer would be false.

Explanation:

prevailing winds blow east-west rather than north-south.

Answer:

A pure substance is a form of matter that has a constant composition and properties that are constant throughout the sample. Mixtures are physical combinations of two or more elements and/or compounds.

Answer:

B. Q > K precipitate will form

Explanation:

The reaction is;

Ba(NO3)2(aq) + Na2CO3(aq) ------> BaCO3(s) + 2NaNO3(aq)

Hence the reaction could form a precipitate of BaCO3.

Number of moles of carbonate ions = 50/1000 * 0.10 M =  5 * 10^-3 moles

Number of moles of Barium ions = 20/1000 * 0.10 M = 2 * 10^-3 moles

Total volume after reaction = 20ml + 50ml = 70 ml or 0.07 L

Molarity Barium ions = 5 * 10^-3 moles/ 0.07 L = 0.07 M

Molarity carbonate ions = 2 * 10^-3 moles/ 0.07 L =0.03 M

Q = [Ba^2+] [CO3^2-] = 0.07 * 0.03 = 2.1 * 10^-3

But K = 2.58  ×  10 ^− 9

We can clearly see that Q>K therefore precipitate will form

Answer:Pierre Curie

Explanation:

Pierre Curie

Answer:

explain cash of the following terms in agricultural

Answer:

The cell potential for the given galvanic cell is 0.188 V.

Explanation:

To calculate the cell potential, we can use the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard cell potential, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25°C = 298 K), n is the number of moles of electrons transferred (in this case, n = 2), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

First, we need to write the half-reactions and their standard reduction potentials:

Sn4+(aq) + 2e- → Sn2+(aq) E°red = 0.15 V

Fe3+(aq) + e- → Fe2+(aq) E°red = 0.77 V

The overall reaction is the sum of the half-reactions:

Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)

The reaction quotient Q can be expressed as:

Q = [Sn4+][Fe2+]^2 / [Sn2+][Fe3+]^2

Substituting the given concentrations, we get:

Q = (0.00655)(0.01139)^2 / (0.0624)(0.0437)^2 = 0.209

Now we can calculate the cell potential:

Ecell = 0.15 V + 0.0592 V log([Fe2+]^2/[Fe3+]) + 0.0592 V log([Sn4+]/[Sn2+])

= 0.15 V + 0.0592 V log(0.01139^2/0.0437^2) + 0.0592 V log(0.00655/0.0624)

= 0.188 V

Therefore, the cell potential for the given galvanic cell is 0.188 V.

The cell potential for the given galvanic cell in which the given reaction occurs at 25 °C is 0.188 V.

To find the cell potential, we take the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

In which R is the gas constant (8.314 J/mol·K) and E° cell is the standard cell potential.

T temperature in Kelvin (25°C = 298 K), and n is the number of moles of electrons transferred (n = 2), Q is the reaction quotient and F is the Faraday constant (96,485 C/mol).

Firstly, write the half-reactions and then their standard reduction potentials:

Sn⁴⁺(aq) + 2e⁻  → Sn²⁺(aq) E°red = 0.15 V

Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E°red = 0.77 V

The overall reaction is the sum of the half-reactions:

Sn²⁺(aq) + 2Fe³⁺(aq) → Sn⁴⁺(aq) + 2Fe²⁺(aq)

The Q reaction quotient can be written as:

Q = [Sn⁴⁺][Fe²⁺]² ÷ [Sn²⁺][Fe²⁺]²

Substituting the given concentrations, we observe:

Q = (0.00655)(0.01139)² ÷ (0.0624)(0.0437)² = 0.209

Next, we can find the cell potential:

Ecell = 0.15 V + 0.0592 V log([Fe²⁺]²/[Fe³⁺]) + 0.0592 V log([Sn⁴⁺]/[Sn²⁺])

= 0.15 V + 0.0592 V log(0.01139²÷0.0437²) + 0.0592 V log(0.00655÷0.0624)

= 0.188 V

Thus, the cell potential for the given galvanic cell is 0.188 V.

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The statements that are true are:

The Ni(s)/Ni2+ (aq) serves as the cathode.

The electrons flow from the Ni(s) to the Sn(s).

The electrons flow from the cathode to the anode.

In an electrolytic cell, oxidation occurs at the anode.

In an electrolytic cell, the electrode with the lower reduction potential, or the cathode, is the one from which electrons flow. Since Ni(s)/Ni2+ (aq) has a reduction potential of -0.25V, it is the cathode of the cell. The electrons flow from the cathode to the anode, which has a higher reduction potential. In this case, the anode is Sn(s)/Sn2+ (aq), which has a reduction potential of -0.14V. The electrons flow from the Ni(s) to the Sn(s) so that oxidation occurs at the anode, as electrons are removed from the system. A salt bridge is required to complete the circuit, as it allows for the flow of ions between the two solutions. A power supply is also required, but the voltage is not specified in the question. Thus, the statement that a minimum of 0.10V is required from a power supply for electrons to flow in the electrolytic cell is false.

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Answer:

F2

Explanation:

F2 is the only element with only 1 element and eveything else is formed by 2 non-metals

Answer:d I believe

Explanation:

Answer:

C

Explanation:

Hope it helps!

The reaction's equilibrium constant is 7.03×10¹².

The equilibrium constant for the reaction can be found using the following expression:

K = [Pb(OH)₃⁻][I⁻]²/ [PbI₂][OH⁻]³

The values of Ksp and Kf can be used to find the concentrations of the species involved in the reaction. Since Pb(OH)₃⁻ is a product of the reaction, its concentration can be expressed in terms of [PbI₂] and [OH⁻]:

[Pb(OH)₃⁻] = Kf[Pb²⁺][OH⁻]³ / (1 + Kf[Pb²⁺])

Since the reaction involves the dissolution of PbI₂, the initial concentration of PbI₂ can be assumed to be equal to its solubility product:

[PbI₂] = Ksp^(1/2)

Substituting these expressions into the equilibrium constant expression:

K = (Kf / Ksp^(3/2)) [OH⁻]³ / (1 + Kf[Pb²⁺])

Using the given values for Ksp and Kf:

K = (8×10¹³ / (8.70×10⁻⁹)^(3/2)) [OH⁻]³ / (1 + 8×10¹³ (Ksp^(1/2) / [OH⁻]))

Simplifying this expression:

K = 7.03×10¹² [OH⁻]³ / ([OH⁻] + 6.51×10⁻⁵)

Therefore, the equilibrium constant for the reaction is 7.03×10¹².

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