in the sum of 54.34 45.66, the number of significant figures is

2100 g Fe₂(SO₄)₃

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Stoichiometry

Atomic Structure

Step 1: Define

5.26 mol Fe₂(SO₄)₃

Step 2: Identify Conversions

Molar Mass of Fe - 55.85 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe₂(SO₄)₃ - 2(55.85) + 3(32.07) + 12(16.00) = 399.91 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2103.53 g Fe₂(SO₄)₃ ≈ 2100 g Fe₂(SO₄)₃

Answer:

9*(10+9)

(90+81)

(171) is the answer

The question that will best help the student to classify the material is; "is the material malleable or ductile?"

Generally, materials can be classified as metals or non metals. There are properties that are particular to metals and there are properties that are particular to nonmetals and these properties can be used to identify each one of the materials.

The question that will best help the student to classify the material is; "is the material malleable or ductile?" These metallic properties.

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Missing parts;

A student is trying to classify an unidentified, solid gray material as a metal or a nonmetal. Which question will best help the student classify the material? A. Is the material malleable or ductile? B. Does the material feel hard to the touch? C. Will the material float in water? D. Does the material feel rough or smooth?

Answer:

The phenotypic percentage of having yellow or green seeds is 50% for having either of the two colours

Explanation:

The crossing to determine the offsprings is shown in the image attached where we have two green (Yy) seeds and two yellow (yy) seeds as offsprings.

Thus, the phenotypic percentage of having yellow or green seeds is 50% for having either of the two colours.

Note, the dominant allele is "Y" while the recessive allele is "y". Thus, Yy would produce a yellow colour while yy would produce a green colour (as both mentioned in the question).

Also note that phenotype describes the outward properties/characteristics of an individual.

We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

Assuming that the number of moles of gas and the pressure are constant, we can write:

V/T = constant

This means that the product of volume and temperature is constant as long as the pressure and the number of moles of gas are held constant.

We can use this relationship to solve the problem.

Let V1 be the initial volume of the gas (150 cm³) and T1 be the initial temperature in Kelvin. Let V2 be the final volume of the gas (250 cm³) and T2 be the final temperature in Kelvin.

We can write:

V1/T1 = V2/T2

Solving for T2:

T2 = (V2/V1) * T1

The cubic expansitivity of gas at constant pressure is given by:

1/273K-1

We can convert the initial temperature from Celsius to Kelvin:

T1 = (0°C + 273.15) K = 273.15 K

Plugging in the values, we get:

T2 = (250 cm³ / 150 cm³) * 273.15 K

T2 = 455.25 K

Therefore, the temperature of the gas when its volume is 250 cm³ is 455.25 K (182.1°C).

Approximately 3.72 liters of C2H2 are required to produce 8 L of CO2 at STP.

We can use stoichiometry to determine the amount of C2H2 required to produce 8 L of CO2 at STP.

According to the balanced equation, 2 moles of C2H2 react with 5 moles of O2 to produce 4 moles of CO2 and 2 moles of H2O. This means that the ratio of C2H2 to CO2 is 2:4 or 1:2.

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 8 L of CO2 corresponds to 8/22.4 ≈ 0.357 moles of CO2.

Since the ratio of C2H2 to CO2 is 1:2, we need half as many moles of C2H2 as moles of CO2:

0.357 moles of CO2 × (1 mole of C2H2 / 2 moles of CO2) = 0.179 moles of C2H2

Now we can use the ideal gas law to convert moles of C2H2 to liters at STP:

PV = nRT

where P is the pressure (1 atm), V is the volume (unknown), n is the number of moles (0.179 moles), R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin (273 K at STP).

Solving for V, we get:

V = nRT/P = (0.179 mol)(0.0821 L·atm/K·mol)(273 K)/(1 atm) ≈ 3.72 L

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Answer:

yes males can give birth

Answer: things arnent always as they seem

Explanation:

After 93.8 hours have passed, 11.6 mg of gallium-67 remains from 46.4 mg of this gallium-67.


The half-life of gallium-67 is 78.2 hours, which means that every 78.2 hours, half of the original amount of the isotope will decay. Using this information, we can determine how much gallium-67 will remain after 93.8 hours have passed.
First, we need to determine how many half-lives have passed in 93.8 hours. We can do this by dividing 93.8 hours by the half-life of gallium-67:

93.8 hours ÷ 78.2 hours/half-life = 1.2 half-lives

This means that 1.2 half-lives have passed, and we can calculate how much gallium-67 remains using the formula:
Amount remaining = (Initial amount) x (1/2)^(number of half-lives)

Plugging in the values we have:

Amount remaining = (46.4 mg) x (1/2)^(1.2) = 11.6 mg

Therefore, after 93.8 hours have passed, 11.6 mg of gallium-67 remains.

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Answer:

98L is the volume of the gas

Explanation:

To solve this question we need to find  the moles of O2 + moles HCl. With these moles and PV = nRT we can find the volume as follows:

Moles O2: 3.0mol

Moles HCl:

50.0g * (1mol/36.5g) = 1.37 moles HCl

Total moles:

3.0mol O2 + 1.37 mol HCl = 4.37 moles

PV = nRT

V = nRT/P

Where V is volume in L, n are moles = 4.37mol, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP, P = 1atm at STP

V = 4.37mol*0.082atmL/molK*273.15K / 1atm

V = 98L is the volume of the gas

Option B. I and IV. ii. the stationary phase is non-polar and iv. the mobile phase is solventless and more polar than silica-gel.

The stationary phase is made up of silica-gel and it is polar and the mobile phase is less polar than this. Due to this difference in polarity, the organic compounds get separated with mobility with the mobile phase.

The desk-bound segment in chromatography is the one which does no longer circulate with the sample while the mobile section in chromatography is one that moves with the sample.​ ​ instance: ​ In Paper Chromatography:​ The paper strip acts as a desk-bound phase at the same time as the solvent act as a "mobile section" in paper chromatography. ​

Skinny layer chromatography is accomplished precisely as it says - the use of a skinny, uniform layer of silica gel or alumina covered onto a chunk of glass, metallic or rigid plastic. The silica gel (or the alumina) is the stationary segment.

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Disclaimer:- your question is incomplete, please see below for the complete question.

Which of the following statements about the stationary and mobile phases in thin-layer chromatography is (are) true? i. the stationary phase is made of silica gel. ii. the stationary phase is a non-polar solvent. iii. the mobile phase is water. iv. the mobile phase is solventless and more polar than silica-gel.

A. I and II

B. I and IV.

C. II and IV.

D. II and III.

Answer:

There is no picture but the one that is in motion in the picture has kinetic energy.

Explanation:

Kinetic energy is the energy of motion.

hope this helped!

Answer:

How many gallons of water are on earth?

Why does salt water make you sticky and burn our eyes so bad?

Explanation:

These are just a couple questions I came up with just now

I hope that this helps

The compounds ranked based on their relative Bronsted acidities from strongest to weakest are as follows:

1. H-I (Hydrogen iodide)

2. H-CH3 (Methyl radical)

3. H-OH (Hydroxide ion)

4. H-NH2 (Ammonia)

5. H-F (Hydrogen fluoride)

Bronsted acidities can be determined by analyzing the stability of the corresponding conjugate bases. A stronger acid will have a more stable conjugate base. Here is the explanation for the ranking:

1. H-I: Hydrogen iodide (HI) is a strong acid because iodide ion (I-) is a stable conjugate base. Iodide ion is large and can effectively disperse negative charge, leading to stability.

2. H-CH3: Methyl radical (CH3) is weaker than HI but stronger than the remaining compounds. It is a stable radical and has resonance structures that stabilize its conjugate base.

3. H-OH: Hydroxide ion (OH-) is less acidic than HI and CH3. It forms a stable conjugate base, but it is not as stable as iodide ion or the methyl radical.

4. H-NH2: Ammonia (NH3) is weaker than the previous compounds. The lone pair on the nitrogen atom can be donated to accept a proton, making NH2- a relatively unstable conjugate base.

5. H-F: Hydrogen fluoride (HF) is the weakest acid among the given compounds. The fluoride ion (F-) is a relatively strong base, and its conjugate acid, HF, is a weaker acid compared to the others.

The ranking of the given compounds based on their relative Bronsted acidities, from strongest to weakest, is H-I, H-CH3, H-OH, H-NH2, and H-F. This ranking is determined by analyzing the stability of their respective conjugate bases, with stronger acids having more stable conjugate bases.

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Answer:

Positive.

Explanation:

The slope is positive and increasing, thus meaning there is a positive correlation between time spent studying each day and performance.

Answer:

21.18 gm of H2

Explanation:

First you need the reaction equation:

N2 + 3H2 ===> 2 NH3

 you can see that 3/2  as many moles of H2 are required as are NH3

   this would be 10.59 moles of H2        (7.06 * 3/2)

      one mole of H2 = 2 gm

         10.59 * 2 = 21.18 gm

The mass defect in Fe-56 is equal to 0.52823 amu.

The mass defect can be defined as the difference between the actual atomic mass and the theoretical mass calculated by addition of the mass of protons and neutrons in the nucleus.

The actual atomic mass is always less than the predicted mass determined by adding the masses of nucleons. This additional mass is due to the binding energy that is released when a nucleus is formed.

\(\triangle M = (Zm_p + Nm_n) - M_A\)

Given the mass of the proton = 1.00728 amu

The given mass of a neutron = 1.008665 amu

The given mass of an electron = \(0.00055 amu\)

The mass of an Fe-56 nucleus = 55.921 amu

The mass defect for Fe-56 can be calculated as:

\(\triangle M = (26\times 1.00728 + 30 \times 1.008665) - 55.921\)

ΔM = 0.52823 amu

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Answer:

steel solder brass Peter duralumin bronze and amalgams

Answer:

I hope it help you,

To prepare 1.00 L of pH = 12.00 buffer with a total concentration of 0.500 M of dimethylamine and dimethylammonium chloride, you will need  0.495 mol of each component.

The Henderson-Hasselbalch equation for a weak base and its conjugate acid is:

pH = pKa + log([conjugate base]/[acid])

At pH 12.00 and with a pKa of 10.73 for dimethylamine, the conjugate base/acid ratio is:

100 = 10.73 + log([conjugate base]/[acid])

[conjugate base]/[acid] = 100/\(10^(^1^0^.^7^3^)\)

[conjugate base]/[acid] = 99.54

Since the total concentration of the two components is 0.500 M, we have:

[conjugate base] + [acid] = 0.500 M

Substituting [conjugate base]/[acid] = 99.54, we can solve for the concentrations of each component:

[acid] = 0.500 M/(1 + 99.54) = 0.00498 M

[conjugate base] = 0.500 M - 0.00498 M = 0.495 M

The amount in moles of each component required to make 1.00 L of the buffer is then:

moles of dimethylamine = 0.00498 mol/L x 1.00 L = 0.00498 mol

moles of dimethylammonium chloride = 0.495 mol/L x 1.00 L = 0.495 mol

Since we need equal amounts of each component in the buffer, we will use the smaller amount, which is 0.00498 mol. Therefore, we need:

mass of dimethylamine = 0.00498 mol x 45.08 g/mol = 0.224 g

mass of dimethylammonium chloride = 0.00498 mol x 81.05 g/mol = 0.404 g

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Answer:

Sand dunes are created when wind deposits sand on top of each other until a small mound starts to form. Once that first mound forms, sand piles up on the windward side more and more until the edge of the dune collapses under its own weight.

a. The half-life of radon-222 is 3.82 days. b. It will take approximately 11.46 days for the sample to decay to 20% of its original amount.

(a) To find the half-life of radon-222, we can use the formula:

\(N = N0 * (1/2)^{(t/T)}\)

where:

\(N = amount\ remaining\ after\ time\ t\\N0 = initial\ amount\\T = half\ -life\)

We know that after 3 days, the amount remaining is 58% of the original amount, so N/N0 = 0.58 and t = 3 days. Substituting these values:

\(0.58 = (1/2)^(3/T)\)

Taking the natural logarithm of both sides:

\(ln(0.58) = ln(1/2)^{(3/T)} \\ln(0.58) = -(3/T) * ln(2)\\T = -(3/ln(2)) * ln(0.58)\\T = 3.82 days\)

(b) To find how long it will take the sample to decay to 20% of its original amount: \(N = N0 * (1/2)^{(t/T)}\)

We want to find the time t for which N/N0 = 0.20. Substituting this value and T = 3.82 days into the formula gives:

\(0.20 = (1/2)^{(t/3.82)}\)

Taking the natural logarithm of both sides:

\(ln(0.20) = (t/3.82) * ln(1/2) \\t = -(3.82/ln(1/2)) * ln(0.20)\)

\(t = 11.46 days\)

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i think that the answer is D

The largest atom or ion among the options given is option D, Ca2+.

What are atoms and ions?

An atom is the basic unit of matter that makes up all elements. An atom is composed of a nucleus, which contains protons and neutrons, and electrons, which occupy energy levels or orbitals surrounding the nucleus.

An ion is an atom or molecule that has gained or lost one or more electrons. When an atom loses one or more electrons, it becomes a positive ion, or cation. When an atom gains one or more electrons, it becomes a negative ion, or anion. Ions are formed by the transfer of electrons from one atom to another.

The size of an atom or ion is determined by the number of electrons it has in its outermost energy level, or valence shell. The more electrons an atom or ion has in its valence shell, the larger it will be. The valence electron number (VNE) of an element is a measure of how many electrons it has in its valence shell.

Option A, K, is the symbol for potassium, which has 19 electrons, so it has a VNE of 1.

Option B, K+, is a potassium ion that has lost one electron, so it has 18 electrons and a VNE of 0.

Option C, Ca, is the symbol for calcium, which has 20 electrons, so it has a VNE of 2.

Option D, Ca2+, is a calcium ion that has lost two electrons, so it has 18 electrons and a VNE of 0.

Option E, Li, is the symbol for lithium, which has 3 electrons, so it has a VNE of 1.

Since Ca2+ has less electrons than K, Ca, and Li, it has a smaller electron cloud, which makes it the largest among the options given.

Hence, the largest atom or ion among the options given is option D, Ca2+.

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Microwave radiation is a form of electromagnetic energy that can penetrate the reaction mixture and directly interact with the molecules, causing them to vibrate and generate heat.

In the context of Fischer Esterification, the microwave radiation can excite the reactant molecules and provide sufficient energy to break the esterification reaction's activation barrier. The heat generated by microwave radiation allows for faster reaction rates and improved yields.

Microwave-promoted reactions refer to those that employ microwave radiation to increase the reaction rate or enhance the yield of the desired product. By using microwave radiation, the reaction can be completed in a shorter time, which can increase the efficiency and throughput of the process.

Microwave-promoted reactions can also lead to higher product purity and selectivity, especially in cases where traditional heating methods may result in side reactions or product degradation. In summary, microwave-promoted reactions can provide a more efficient and sustainable method of chemical synthesis.

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In an endothermic reaction, the energy level increases as the reaction proceeds starting from the reactants and ending with the products.

An endothermic reaction is a chemical reaction that absorbs energy from its surroundings in the form of heat. This heat is generally transferred from the surroundings to the reaction mixture, and as a result, the temperature of the mixture decreases. An endothermic reaction is often depicted with a positive change in enthalpy.

The reaction absorbs heat, and the energy of the system is increased. In the forward reaction, reactants absorb heat and increase in energy, which is manifested as an increase in the reaction vessel's temperature. The molecules then begin to vibrate more quickly and vigorously.

They then overcome the activation energy barrier and form the products.The enthalpy of the products is more than the enthalpy of the reactants, as energy was absorbed in the reaction. As a result, the energy level increases as the reaction proceeds starting from the reactants and ending with the products. Therefore, the correct option is 2) The energy level increases.

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In a voltaic cell with a standard hydrogen electrode (SHE) and a Cu/Cu2+ half-cell, we can determine the Cu2+ concentration when the cell potential (E_cell) is 0.060 V. The SHE is assigned a potential of 0 V, and for the Cu/Cu2+ half-cell, the standard reduction potential (E°) is 0.34 V. To calculate the Cu2+ concentration, we will use the Nernst equation:
E_cell = E° - (RT/nF) * ln(Q)
Now, solve for Q, which represents [Cu2+]/[H+]^2. Since [H+] in SHE is 1 M, Q equals [Cu2+]. After solving for Q, you'll find the concentration of Cu2+ in the Cu/Cu2+ half-cell.

In order to calculate [Cu2+] in the given voltaic cell, we need to use the Nernst equation:
Ecell = E°cell - (RT/nF)ln(Q)
Where Ecell is the cell potential, E°cell is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the cell reaction, F is Faraday's constant, and Q is the reaction quotient.
Since the half-cell with the standard hydrogen electrode is the reference half-cell, its standard reduction potential is defined as 0 V. Therefore, the standard cell potential for the given cell can be calculated as follows:
E°cell = E°Cu/Cu2+ - E°H+/H2
Where E°Cu/Cu2+ is the standard reduction potential for the Cu/Cu2+ half-cell, which is 0.34 V. Thus:
E°cell = 0.34 V - 0 V = 0.34 V
We can rearrange the Nernst equation to solve for [Cu2+]:
ln([Cu2+]/[Cu]) = (nF/RT)(E°cell - Ecell)
Substituting the given values:
ln([Cu2+]/[Cu]) = (2)(96485 C/mol)/(8.314 J/K/mol)(298 K)(0.34 V - 0.060 V)
Solving for [Cu2+]:
[Cu2+] = [Cu]e^(nF/RT)(E°cell - Ecell)
[Cu2+] = [Cu]e^(2)(96485 C/mol)/(8.314 J/K/mol)(298 K)(0.28 V)
Assuming that [Cu] remains constant at a concentration of 1 M:
[Cu2+] = 1 M e^(-0.0097) = 0.990 M
Therefore, [Cu2+] in the given voltaic cell is 0.990 M when Ecell is 0.060 V.

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The hydrogen can be produced from water using renewable energy sources, which makes it more sustainable.

If we based our energy on hydrogen, two major economic or global problems that could be alleviated are:

1. Climate change: This is a global issue that requires an immediate response. The world needs to move away from carbon-emitting fossil fuels. Burning hydrogen fuel emits only water and does not release greenhouse gases. If the world shifts to hydrogen fuel, it will reduce carbon emissions and help to slow down climate change.

2. Dependence on Oil: Most countries are dependent on oil. The price of oil is volatile, and the demand and supply fluctuate due to political, economic, and weather events. This dependence on oil is a major economic challenge for many countries. If we based our energy on hydrogen, we could reduce our dependence on oil and decrease oil imports, which could significantly improve the economy of countries that do not produce oil.

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Answer:

B. melting of ice caps

Explanation:

Answer:

livermorium

Explanation:

The atom with the largest number of neutrons is a tie between livermorium and tennessine. Each of these atoms contain 177 neutrons.