draw the basic structure of an amino acid. how does the structure of an amino acid contribute to its function? use the terms r group, folding, peptide bond, polarity, ribosome, and carbon.

Aldehydes that contain at least one alpha hydrogen atom can undergo Aldol condensation and Cannizzaro reaction.

As per the question given,  

In Aldol condensation, two aldehyde or ketone molecules react in the presence of a base to form a beta-hydroxy carbonyl compound. The reaction involves the removal of a water molecule from the alpha-carbon and carbonyl group of two aldehydes or ketones. This reaction produces Aldol, which can undergo dehydration to form an alpha-beta-unsaturated carbonyl compound.

On the other hand, in the Cannizzaro reaction, an aldehyde molecule undergoes self-oxidation and reduction in the presence of a strong base to give an alcohol and a carboxylic acid. The reaction involves the loss of a hydrogen atom from the aldehyde, which gets reduced to an alcohol, and the reduction of the other aldehyde molecule to a carboxylic acid.

Therefore, aldehydes containing at least one alpha hydrogen atom can undergo Aldol condensation and Cannizzaro reaction.

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the answer is ultraviolet

Answer:

microwave

Explanation:

your gonna have one crispy plant if you put it in the microwave XD

In the given lab write-up, the third possibility for the mechanism of the rate-determining step was confirmed, where the rate-determining step involves an iodide ion and a persulfate ion coming together.

If the first mechanism were correct, where the rate-determining step has two iodide ions coming together, the rate of the reaction would be second order with respect to the iodide ion concentration, and doubling the iodide ion concentration would increase the rate by a factor of four. If the first mechanism were correct, where the rate-determining step involves a persulfate ion decomposing, the rate of the reaction would be first order with respect to the persulfate ion concentration, and doubling the persulfate ion concentration would double the rate.

However, since the third mechanism was confirmed, where the rate-determining step has an iodide ion and a persulfate ion coming together, the rate of the reaction is second order overall, and doubling either the iodide or persulfate ion concentration would increase the rate by a factor of four.

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Full Question: your lab write-up, three possibilities for the mechanism of the rate-determining step were listed. 1. The rate-determining step has two iodide ions coming together. 2. The rate-determining step involves a persulfate ion decomposing. 3. The rate-determining step has an iodide ion and a persulfate ion coming together. Which mechanism did your experiment confirm? the third. (a) If the first mechanism is correct, what should happen to the rate if the concentration of iodide ion is doubled and other concentrations are held constant? (b) If the first mechanism is correct, what should happen to the rate if the concentration of persulfate ion is doubled and other concentrations are held constant? (c) If the second mechanism is correct, what should happen to the rate if the concentration of iodide ion is doubled and other concentrations are held constant? (d) If the second mechanism is correct, what should happen to the rate if the concentration of persulfate ion is doubled and other concentrations are held constant?

.

Answer:

To solve this problem, we'll need to use stoichiometry and the molar ratios from the balanced chemical equation. Here's how you can calculate the values:

(a) Mass of the lead nitrate sample:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 1 mole of oxygen gas (O2). We know that the volume of oxygen gas collected is 112 cm³, which is equal to 112/1000 = 0.112 dm³ (converting cm³ to dm³).

According to the molar volume of gas at STP (22.4 dm³), 1 mole of any gas occupies 22.4 dm³ at STP. Therefore, the number of moles of oxygen gas can be calculated as:

moles of O2 = volume of O2 / molar volume at STP

moles of O2 = 0.112 dm³ / 22.4 dm³/mol = 0.005 mol

Since 2 moles of lead nitrate produce 1 mole of oxygen gas, we can determine the number of moles of lead nitrate as:

moles of Pb(NO3)2 = 2 * moles of O2

moles of Pb(NO3)2 = 2 * 0.005 mol = 0.01 mol

To calculate the mass of the lead nitrate sample, we'll use its molar mass:

mass of Pb(NO3)2 = moles of Pb(NO3)2 * molar mass of Pb(NO3)2

mass of Pb(NO3)2 = 0.01 mol * (207 g/mol + 2 * 14 g/mol + 6 * 16 g/mol)

mass of Pb(NO3)2 = 0.01 mol * 331 g/mol

mass of Pb(NO3)2 = 3.31 g

Therefore, the mass of the lead nitrate sample is 3.31 grams.

(b) Mass of lead(II) oxide produced:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 2 moles of lead(II) oxide (PbO). So, the number of moles of PbO produced is equal to the number of moles of Pb(NO3)2.

mass of PbO = moles of PbO * molar mass of PbO

mass of PbO = 0.01 mol * (207 g/mol + 16 g/mol)

mass of PbO = 0.01 mol * 223 g/mol

mass of PbO = 2.23 g

Therefore, the mass of lead(II) oxide produced is 2.23 grams.

(c) Volume of nitrogen dioxide gas produced at STP:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 4 moles of nitrogen dioxide gas (NO2). So, the number of moles of NO2 produced is twice the number of moles of Pb(NO3)2.

moles of NO2 = 2 * moles of Pb(NO3)2

moles of NO2 = 2 * 0.01 mol = 0.02 mol

Using the molar volume of gas at STP, we can calculate the volume of nitrogen dioxide gas:

volume of NO2 = moles of NO2 * molar volume at STP

volume of NO2 = 0.02 mol * 22.4 dm³/mol = 0.448 dm³

Therefore, the volume of nitrogen dioxide gas

Explanation:

Matter is the basic unit of all things in the world, whether living or nonliving. It is not only useful in chemistry, but also to other fields of science like physics. Physics deals with motion, so it has something to do with matte especially the movement of molecules for solids and fluids.

To make 450 mL of a 0.500 M Zn²⁺ solution, the chemist needs to mix 108 mL of 1.45 M ZnSO₄ and 342 mL of 0.200 M ZnCl₂. The ion with the lowest concentration in the final solution is Zn²⁺.

Let's assume x mL of the 1.45 M ZnSO₄ solution is needed.

The concentration of Zn²⁺ in the ZnSO₄ solution is 1.45 M, so the amount of Zn²⁺ in this solution is (1.45 M)(x mL) = 1.45x mmol.

The remaining volume of the solution required is (450 - x) mL.

To obtain a 0.500 M Zn²⁺ concentration, the amount of Zn²⁺ needed in the ZnCl₂ solution is (0.500 M)(450 - x) mL = 0.5(450 - x) mmol.

Since the volumes are additive, the total amount of Zn²⁺ required is 1.45x mmol + 0.5(450 - x) mmol.

Setting the total amount equal to the desired concentration:

1.45x + 0.5(450 - x) = 0.500(450)

1.45x + 225 - 0.5x = 225

0.95x = 0

x = 0

Thus, 0 mL of the 1.45 M ZnSO₄ solution is needed.

To find the volume of the 0.200 M ZnCl₂ solution:

Volume of ZnCl₂ solution = Total volume - Volume of ZnSO₄ solution

= 450 mL - 0 mL

= 450 mL

Therefore, 450 mL of the 0.200 M ZnCl₂ solution is needed.

Since the concentration of Zn²⁺ is higher in the ZnSO₄ solution (1.45 M) than in the ZnCl₂ solution (0.200 M), Zn²⁺ has the lowest concentration in the final solution.

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the concentrations: Ka = ((1.82 × 10^(-5))(1.82 × 10^(-5)))/(0.45) ≈ 7.34 × 10^(-11) then the Ka of the weak acid HX is approximately 7.34 × 10^(-11).

To determine the Ka of the weak acid HX, we'll follow these steps:

1. Identify the salt (NaX) and its corresponding weak acid (HX) and strong base (NaOH).
2. Calculate the pOH of the solution.
3. Determine the concentration of OH- ions.
4. Find the concentration of X- ions, and HX.
5. Set up an equilibrium expression and solve for Ka.

Step 1: The salt given is NaX, which comes from the weak acid HX and the strong base NaOH.

Step 2: Since the pH of the solution is 9.26, we can calculate the pOH using the relationship pH + pOH = 14.
pOH = 14 - pH = 14 - 9.26 = 4.74

Step 3: To find the concentration of OH- ions, we'll use the relationship pOH = -log[OH-].
Rearrange the equation and solve for [OH-]: [OH-] = 10^(-pOH) = 10^(-4.74) ≈ 1.82 × 10^(-5) M

Step 4: The concentration of X- ions in the salt solution is 0.45 M (given), and since NaOH is a strong base, the reaction with water is complete. Thus, [X-] after the reaction with water will be 0.45 - [OH-] ≈ 0.45 M, and [HX] ≈ [OH-] ≈ 1.82 × 10^(-5) M.

Step 5: Set up the equilibrium expression for Ka: Ka = ([HX][OH-])/([X-]).
Substitute the concentrations: Ka = ((1.82 × 10^(-5))(1.82 × 10^(-5)))/(0.45) ≈ 7.34 × 10^(-11)

So, the Ka of the weak acid HX is approximately 7.34 × 10^(-11).

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The balanced chemical equation for the reaction between magnesium and oxygen to form magnesium oxide is:

2 Mg + O2 → 2 MgO

In the reaction, magnesium (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO). To balance the equation, we need to ensure that the number of atoms on both sides of the equation is the same.

In this case, we have two magnesium atoms on the left side and two magnesium atoms on the right side, which are already balanced. However, we have two oxygen atoms on the right side (in the form of O2), so we need to balance the equation by placing a coefficient of 2 in front of MgO on the left side.

After balancing, we have two magnesium atoms and two oxygen atoms on both sides of the equation:

2 Mg + O2 → 2 MgO

Therefore,

The balanced chemical equation for the reaction between magnesium and oxygen to form magnesium oxide is 2 Mg + O2 → 2 MgO.

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The new volume of the balloon at the top of Pikes Peak is approximately 5.94 liters.

To solve this problem, we can use the combined gas law equation, which relates the initial and final states of a gas. The equation is given as:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Given:

P1 = 566.6 torr (initial pressure at 8500 ft)

V1 = 8.6 L (initial volume)

T1 = 26.1°C (initial temperature)

P2 = 400 torr (pressure at the top of Pikes Peak)

T2 = 1.8°C (temperature at the top of Pikes Peak)

We need to solve for V2, the new volume of the balloon at the top of Pikes Peak.

First, let's convert the temperatures to Kelvin:

T1 = 26.1°C + 273.15 = 299.25 K

T2 = 1.8°C + 273.15 = 275.95 K

Now we can plug the values into the equation:

(566.6 torr * 8.6 L) / (299.25 K) = (400 torr * V2) / (275.95 K)

To solve for V2, we can cross-multiply and then divide:

(566.6 torr * 8.6 L * 275.95 K) / (400 torr * 299.25 K) = V2

Calculating this expression, we find:

V2 ≈ 5.94 L

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The concentration of Ca2+ ions inside the cell is typically much lower than outside the cell, typically around 0.0001-0.001 mM.

This is due to the activity of ion pumps and channels that work to maintain this concentration gradient across the cell membrane. Alternatively, the concentration of Ca2+ ions inside a cell is typically lower than outside. While the concentration outside the cell is 2.0 mM, the concentration inside the cell is usually around 100 nM. This difference in concentration is maintained by various cellular mechanisms such as calcium pumps and ion channels.

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Answer:

The formula is CI 3 I. Hope this helps!

Answer:

ICl3.

Explanation: trust me

1. The new volume would be approximately 0.082 L. for 2, The new pressure would be approximately 135.7 kPa. for 3, The new temperature would be approximately 223.3 K. for 4, The new temperature would be approximately 377.3 K. for 5, The new volume would be approximately 0.07 L.

1, Given: P₁ = 202 kPa, V₁ (volume)= 100 mL = 0.1 L, T₁ (temperature )= 25°C = 298 K ,P₂ = 250 kPa ,T₂ = 35°C = 308 K

Using the combined gas law formula, one can rearrange it to solve for V₂: V₂ = (P₁V₁T₂) / (P₂T₁)

V₂ = (202 kPa × 0.1 L × 308 K) / (250 kPa × 298 K)

V₂ ≈ 0.082 L

The new volume would be approximately 0.082 L.

2, Given: V₁ = 150 mL = 0.15 L ,T₁ = 100°C = 373 K ,P₁ = 303.9 kPa, T₂ = 50°C = 323 K ,V₂ = 225 mL = 0.225 L

Again, using the combined gas law formula and solving for P₂: P₂ = (P₁V₁T₂) / (V₂T₁)

P₂ = (303.9 kPa × 0.15 L × 323 K) / (0.225 L × 373 K)

P₂ ≈ 135.7 kPa

The new pressure would be approximately 135.7 kPa.

3, Given: V₁ = 500 mL = 0.5 L, T₁ = 0°C = 273 K (STP) ,P₁ = 101.3 kPa (STP) ,V₂ = 0.5 * V₁ = 0.25 L, P₂ = 2 * P₁ = 2 * 101.3 kPa = 202.6 kPa

To determine the new temperature, we rearrange the formula to solve for T₂:

T₂ = (P₂V₂T₁) / (P₁V₁)

T₂ = (202.6 kPa× 0.25 L ×273 K) / (101.3 kPa ×0.5 L)

T₂ ≈ 223.3 K

The new temperature would be approximately 223.3 K.

4, Given: V₁ = 150 mL = 0.15 L, P₁ = 50 kPa ,T₁ = 10°C = 283 K, P₂ = 200 kPa ,V₂ = 200 mL = 0.2 L

Solving for T₂:

T₂ = (P₂V₂T₁) / (P₁V₁)

T₂ = (200 kPa × 0.2 L × 283 K) / (50 kPa ×0.15 L)

T₂ ≈ 377.3 K

The new temperature would be approximately 377.3 K.

5, Given: V₁ = 150 mL = 0.15 L ,T₁ = 100°C = 373 K, P₁ = 500 kPa ,T₂ = 0°C = 273 K (STP) ,P₂ = 101.3 kPa (STP)

To determine the new volume, we rearrange the formula to solve for V₂:

V₂ = (P₁V₁T₂) / (P₂T₁)

V₂ = (500 kPa × 0.15 L × 273 K) / (101.3 kPa × 373 K)

V₂ ≈ 0.07 L

The new volume would be approximately 0.07 L.

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HCO3^- is best described as ao bicarbonate in.

The bicarbonate ion, HCO3^-, consists of one hydrogen atom (H+), one carbon atom (C), and three oxygen atoms (O) bonded together. It is a polyatomic ion that plays a crucial role in various biological and chemical processes. Bicarbonate ions are commonly found in the body and are involved in maintaining acid-base balance, particularly in blood and cellular environments.

In terms of acidity, HCO3^- can act as a weak acid. It has the ability to donate a proton (H+) in certain chemical reactions, contributing to the regulation of pH levels in the body. However, it is important to note that HCO3^- is primarily known as a bicarbonate ion and is more commonly involved in its role as a base rather than an acid.

In summary, HCO3^- is best described as a bicarbonate ion, which is involved in maintaining acid-base balance and acts as a weak acid in specific reactions describes HCO3^-. a proton donor a bicarbonate ion a weak acid common in the liver

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HCO3^- is known as the bicarbonate ion. It acts as a weak acid or a proton donor, assisting with pH regulation in the blood by buffering acid wastes from metabolic processes. It is also involved in respiratory regulation of acid-base balance.

HCO3^- is known as bicarbonate ion. It can act as a proton donor, thus making it a weak acid. In the body, bicarbonate ions and carbonic acid exist in a 20:1 ratio, helping to maintain blood pH balance. Bicarbonate ions prevent significant changes in blood pH by capturing free ions. During metabolic processes that release acid wastes such as lactic acid, bicarbonate ions help to buffer the acidity. These ions are even involved in respiratory regulation of acid-base balance, as they are crucial to the balance of acids and bases in the body by regulating the blood levels of carbonic acid. The stronger the acidic substance, the more readily it donates protons (H*). In contrast, bicarbonate is a weak base, meaning that it releases only some hydroxyl ions or absorbs only a few protons. Overall, the bicarbonate ion plays a critical role in various biological reactions and maintaining homeostasis.

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Answer:

OB heat

Explanation:

Hope this helps

The average atomic mass of the atom that has the following isotopes in the following abundances: 20 10Z 30% , 19 10Z10% , 18 10Z 60% is 18.7.

The average atomic mass of an isotopic element is the weighted average of the atomic masses of the naturally occurring isotopes of an element.

According to this question, the atom has the following isotopes in the following abundances: 20 10Z [30%] = 30/100 × 20 = 6

19 10Z [10%] = 10/100 × 19 = 1.9

18 10Z [60%] = 60/100 × 18 = 10.8

The sum of these masses is the average atomic mass and is calculated as follows:

6 + 1.9 + 10.8 = 18.7

Therefore, the average atomic mass of the atom that has the following isotopes in the following abundances: 20 10Z 30% , 19 10Z10% , 18 10Z 60% is 18.7.

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The compound is named 2,3-dibromo-2,4-hexadienedioic acid, which consists of a six-membered ring with alternating single and double bonds.

The line-angle formula describes a six-membered ring with alternating single and double bonds. To name the compound, we need to consider the functional groups and substituents present.

The carboxylic acid group, -COOH, is attached to the first vertex of the ring. This group is named as "hexanedioic acid" because it contains six carbon atoms in a linear chain. The prefix "hexa-" indicates the presence of six carbons, and the suffix "-dioic acid" denotes the presence of two carboxylic acid groups.

The bromine atom, represented by "Br," is attached to the third vertex in a clockwise direction. Since there are two bromine atoms present, the prefix "di-" is used. Thus, the compound is named "dibromo-hexanedioic acid."

To specify the positions of the bromine atoms, we start numbering the ring from the vertex where the carboxylic acid group is attached, which is the first vertex. Moving clockwise, the second vertex has a double bond, the third vertex has a bromine atom, and the fourth vertex has a double bond. Therefore, the compound is named as "2,3-dibromo-2,4-hexadienedioic acid." The numbers indicate the positions of the substituents in the ring.

In summary, the compound is named 2,3-dibromo-2,4-hexadienedioic acid, which represents a six-membered ring with alternating single and double bonds, a carboxylic acid group attached to the first vertex, and a bromine atom attached to the third vertex in a clockwise direction.

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Answer:

Thomson's results showed that atoms contain smaller particles, so Dalton's theory had to be modified. Rutherford's results showed that most of an atom is empty space and that the mass is concentrated in the center of the atom.

Explanation:

( like and follow if it helps >W< )

Answer:

The Ritcher scale

Explanation:

It measures the magnitiude of earthquakes on the basis of seismograph oscillations.

Answer:

the Richter scale

Explanation:

The sponge is unsaturated when it is taking in more water. It becomes saturated when it stops taking in water. It is supersaturated when water starts oozing out from the sponge.

A saturated solution contains just as much solute as it normally hold at a particular temperature. An unsaturated solution contains less solute than it can normally hold at a particular temperature. A supersaturated solution contains more solute than it can normally hold at a particular temperature.

We can use a sponge to model these three scenario as follows;

Water continues to enter into the sponge when it is unsaturated, this continues until the sponge becomes saturated with water and takes in no more water. When the sponge becomes supersaturated, water begins to ooze out from the sponge because it can no longer hold more water.

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Answer:

E = 21.22×10⁻⁴⁵ J

Explanation:

Given data:

Frequency of photon = 3.20×10⁻¹¹ s⁻¹

Energy of photon = ?

Solution:

Formula:

E = h×f

h = plancks  constant = 6.63×10⁻³⁴ Js

by putting values,

E = 6.63×10⁻³⁴ Js × 3.20×10⁻¹¹ s⁻¹

E = 21.22×10⁻⁴⁵ J

Answer:

it has

Explanation:

energy cannot be made, its always there, it just can be activated to do something.

hope this helps

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208.29 kg of cryolite will be produced when 16.4 kg of Al₂O₃, 56.4 kg of NaOH, and 56.4 kg of HF react completely. The balanced chemical equation for the reaction between Al₂O₃, NaOH, and HF to produce cryolite is:

2 Al₂O₃+ 6 NaOH + 12 HF → 2 Na₃AlF₆ + 9 H₂O

According to the equation, 2 moles of Al₂O₃ react with 6 moles of NaOH and 12 moles of HF to produce 2 moles of cryolite and 9 moles of water.

To calculate the amount of cryolite produced, we need to first convert the given masses of Al₂O₃ , NaOH, and HF to moles by dividing each mass by their respective molar mass. Then, we can use the mole ratios from the balanced equation to determine the number of moles of cryolite produced.

16.4 kg of Al₂O₃ is equal to 0.1 moles
56.4 kg of NaOH is equal to 1.41 moles
56.4 kg of HF is equal to 1.8 moles

From the balanced equation, we can see that 2 moles of Al₂O₃ reacts to produce 2 moles of cryolite. Therefore, 0.1 moles of Al₂O₃ will produce 0.1 moles of cryolite.

Using the mole ratios from the balanced equation, we find that 1.41 moles of NaOH and 1.8 moles of HF react to produce 2 moles of cryolite. Therefore, the limiting reagent is NaOH, and only 1.41 moles of cryolite will be produced.

Finally, we can convert the number of moles of cryolite to its mass by multiplying it by its molar mass:

1.41 moles of cryolite is equal to 208.29 kg of cryolite.

Therefore, 208.29 kg of cryolite will be produced when 16.4 kg of Al₂O₃, 56.4 kg of NaOH, and 56.4 kg of HF react completely.

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The standard reaction enthalpy for the given reaction is +235.8 kJ/mol.

The standard reaction enthalpy (ΔH°) for the given reaction is determined as follows:

Equation of reaction: 3 Fe₂O₃ (s) → 2 Fe₃O₄ (s) + ½ O₂ (g)

The standard enthalpy of formation values for Fe₂O₃ (s), Fe₃O₄(s), and O₂(g) is used to calculate the standard reaction enthalpy.

ΔH° = [2 × ΔH°f(Fe₂O₃)] + [½ × ΔH°f(O₂)] - [3 × ΔH°f(Fe₃O₄)]

where;

ΔH°f(Fe₂O₃) = -824.2 kJ/mol

ΔH°f(Fe₃O₄) = -1118.4 kJ/mol

ΔH°f(O₂) = 0 kJ/mol

ΔH° = [2 × (-1118.4 kJ/mol)] + [½ × 0 kJ/mol] - [3 × (-824.2 kJ/mol)]

ΔH° = -2236.8 kJ/mol + 0 kJ/mol + 2472.6 kJ/mol

ΔH° = 235.8 kJ/mol

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Answer:

A is the correct answer

Answer:

A. The electrons farthest from the nucleus of the atom

Explanation:

The amount of heat that is needed to change 250.0 grams of water at 80°C to steam at 100°C is 20900 J of heat.

Heat change is the amount of heat that must be added or that is evolved when a particular change occurs in a substance.

The amount of heat required is determined from the formula of heat given below as follows:

Heat change = mass * specific heat capacity * temperature change

Heat required = 250 * 4.18 * (100 - 80)

Hee=at required = 20900 J

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Answer: Diagram 4

Explanation: Diagram 1 represents a state in which particles are closely packed , thus cannot be a representation of gas.

Diagram 2 represents  a state in which particles are loosely packed , thus can be a representation of gas.

Diagram 3 represents  a state in which particles are closely packed , thus can be a representation of a solid.

Diagram 4 represents  a state in which particles are loosely packed , thus can be a representation of gas. It also represents the particles in ionized form. Particles are existing as ions bearing positive and negative charges.

Explanation:

Answer:

D

Explanation:

The concentration in units of mole fraction is calculated to be 1.80 × 10⁻⁵ when the solubility of oxygen gas in water at 40° C is 1.0 mmol/l of solution.

In the 1 Liter of a solution, the solubility of 1.0 mmol/L can be written as:

= 0.001 mol of oxygen.

However, the moles of water in 1L can be calculated as follows;

We know that 1 L of water = 1000 mL of water.

Lets assume that the density of the water is 1.0 g/mL

density = mass/volume

Making mass the subject of the formula:

Mass = density × volume

Mass = 1.0 g/mL× 1000 mL

Mass of water = 1000 g

number of moles = mass(g)/molar mass

Given the number of moles of water are 1000 g/ 18.02 g/mol

Given the number of moles of water are 55.49 moles

Mole Fraction of oxygen = no. of moles/ total no. of moles in the system

Mole Fraction of oxygen= 0.001 moles/55.49 moles

mole fraction of oxygen = 1.80 × 10⁻⁵.

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Answer:

2.0 x 10²³ atoms Au = 65.4g

1.5 x 10²³ atoms Pb = 51.6g

7.9 x 10²¹ atoms U = 3.1g

Explanation:

We want to convert each to grams.

We can use dimensional analysis to do so.

Kindly view the attached image for work.

1.) 2.0 x 10²³ atoms Au

Needed conversions:

Math: (2.0x10²³atoms)/(6.022x10²³atoms) x 196.966g = 65.4g

2.) 1.5 x 10²³ atoms lead

Needed conversions:

Math: (1.5x10²³atoms)/(6.022x10²³atoms) x 207.2g = 51.6g

3.) 7.9 x 10²¹ atoms U

Needed conversions:

Math: (7.9x10²¹atoms)/(6.022x10²³atoms) x 238.029g = 3.1g