20. Which of the following statements regarding subshell filling order for a neutral atom is/are CORRECT? 1. Electrons are assigned to the 4f subshell before they are assigned to the 6s subshell. 2. Electrons are assigned to the 5d subshell before they are assigned to the 6p subshell. 3. Electrons are assigned to the 4p subshell before they are assigned to the 4s subshell. a. 1 only b. 2 only c. 3 only d. 1 and 3 e. 1, 2, and 3

Answers

Answer 1

Answer:

2 only

Explanation:

Electrons are filled in atoms according to the Aufbau principle. Electrons are filled into lower energy orbital before the filling of higher energy orbitals and this sequence must be followed in filling electron orbitals.

The order of arrangement of energy levels may be shown as follows; 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p. This implies that 4f>6s, hence 6s is filled before 4f.

Also, the 6p level > 5d level hence this is the correct option. You must fill the 5d level before you feel the 6p level.


Related Questions

Explain why phosphorus has a low melting point.

Answers

Answer:

Phosphorus has a low melting point because the intramolecular forces holding it together is London Dispersion Forces.

Explanation:

London Dispersion Forces (LDF) are the weakest intramolecular forces. You don't need to break the covalent bonds, but rather the Van Der Waals' Forces. If LDF are the weakest forces, then the melting point is low.

Answer:

because the intramolecular forces holding it together is London Dispersion Forces.

Explanation:

Consider the three statements below. Which numbered response contains all the statements that are true and no false statements?
I. Hydration is a special case of solvation in which the solvent is water.
II. The oxygen end of water molecules is attracted toward Ca2+ ions.
III. The hydrogen end of water molecules is attracted toward Cl- ions.
a) I, II, and III
b) I and II
c) III
d) I
e) II

Answers

Answer:

a) I, II, and III

Explanation:

For the first statement;

Solvation, is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. if the solvent is water, we call this process hydration.

This means the statement is TRUE.

For the second statement;

The negatively-charged side of the water molecules are attracted to  positively-charged ions. In the case of water, the oxygen end is the negatively charged side of water. This means the statement is TRUE.

For the third statement;

The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions. In the case of water, the hydrogen end is the positively charged side of water. This means the statement is TRUE.

Going through the options, we can tell that the correct option is option A.

What causes the electrons to flow through a wire in a lead-acid battery?​

Answers

Answer:

The potential difference between the cathode and anode are set up from the chemical reaction. Inside the battery electrons are pushed by the chemical reaction toward the positive end creating a potential difference. It is this potential difference that drives the electrons through the wire.

How are sodium (Na) and potassium (K) similar?
O
A. They rarely react.
B. They are hard metals.
O O
C. They are highly reactive.
D. They are soft nonmetals.

Answers

Answer:

They are also relatively soft metals: sodium and potassium can be cut with a butter knife.

Answer:A

Explanation:

A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound

Answers

Answer:

The empirical formula is: C₂H₃O

Explanation:

The empirical formula, also known as the “minimum formula”, is the simplest expression to represent a chemical compound and indicates the elements that are present and the minimum integer ratio between its atoms.

The percentage composition is the percentage by mass of each of the elements present in a compound.

Having 100 g of the compound as a base, it is possible to express the percentages in grams. That is, assuming you have 100 g of the compound, you have 53.46 g of  C , 6.98 g of  H , and 39.56 g of  O.

Taking into account the molecular mass of each substance, the number of relative atoms of each chemical element is calculated:

C: [tex]53.46 g *\frac{1 mol}{12.01 g } = 4.45 moles[/tex]

H:[tex]6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles[/tex]

O:[tex]39.56 g *\frac{1 mol}{16g } = 2.47 moles[/tex]

Now you divide each value obtained by the least of them:

C: [tex]\frac{4.45 moles}{2.47 moles}= 1.80[/tex]

H:[tex]\frac{6.91 moles}{2.47 moles}= 2.8[/tex]

O:[tex]\frac{2.47 moles}{2.47 moles}=1[/tex]

Decimals approach the nearest integer, then:

C: 2

H: 3

O: 1

Therefore the empirical formula is: C₂H₃O

Answer: The other guy is wrong. The correct answer is C9H14O5

Explanation:

et the mass percentages from the percent composition represent grams in a total mass of 100g. Use the grams to calculate the number of moles of each atom present.

%C:53.46%molC=53.46gC=53.46gC×1molC12.011gC=4.451molC

%H:6.98%molH=6.98gH=6.98gH×1molH1.008gH=6.924molH

%O:39.56%molO=39.56gO=39.56gO×1molO15.999gO=2.473molO

Divide by the smallest number of moles.

subscriptC=4.451molC2.473molO≈1.800≈95

subscriptH=6.924molH2.473molO≈2.800≈145

 

subscriptO=2.473molO2.473molO=1

Now, multiply each subscript by 5 to achieve whole number subscripts. Therefore, the empirical formula is C9H14O5.

Find the pH. What are the pH values for the following solutions? (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.05 M HCl (d) 0.05 M NaOH

Answers

Answer:

(a) [tex]pH=1[/tex]

(b) [tex]pH=1.3[/tex]

(c) [tex]pH=13[/tex]

(d) [tex]pH=12.7[/tex]

Explanation:

Hello,

In this case, we define the pH in terms of the concentration of hydronium ions as:

[tex]pH=-log([H^+])[/tex]

Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:

(a)

[tex][H^+]=[HCl]=0.1M[/tex]

[tex]pH=-log(0.1)=1[/tex]

(b)

[tex][H^+]=[HCl]=0.05M[/tex]

[tex]pH=-log(0.05)=1.3[/tex]

Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

[tex][OH^-]=[NaOH][/tex]

[tex]pOH=-log([OH^-])[/tex]

[tex]pH=14-pOH[/tex]

Thus, we have:

(b)

[tex]pOH=-log(0.1)=1\\pH=14-1=13[/tex]

(d)

[tex]pOH=-log(0.05)=1.3\\pH=14-1.3=12.7[/tex]

Best regards.

Need help with chemistry question

Answers

Answer:

See explanation

Explanation:

In this case, we have to check two variables:

1) The leaving group

2) The carbon bonded to the leaving group.

Let's check one by one:

2-chloro-3-methylbutane

In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a secondary substrate.

1-phenylpropan-1-ol

In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a secondary substrate.

(E)-pent-3-en-2-yl 4-methylbenzenesulfonate

In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an allylic substrate.

3a-bromooctahydro-1H-indene

In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a tertiary substrate.

1-iodo-3-methylbutane

In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a primary substrate.

See figure 1

I hope it helps!

Once the chromatographic column has been prepared, why is it important to allow the level of the liquid in the column to drop the level of the alumina before applying the solution of the compound to be separated?

Answers

Answer: Chromatography is a technique which is used for the separation of components present in a mixture into sub-components.

Explanation:

The solvent or liquid used for the separation of the chromatographic mixture should lie below the spot point where the mixture is loaded in an alumina column. This is done to prevent the air bubbles formations, which can lead to poor sanitation. Also this is done to prevent the mixing of the mixture to be separated with the solvent instead of running with the solvent via capillary action.

Consider the equilibrium reaction. 4A+B↽−−⇀3C After multiplying the reaction by a factor of 2, what is the new equilibrium equation?

Answers

8A+2B——> 6C
since you multiply by a factor of 2 you do that to each letter
4*2=8
1*2=2
3*2=6

The nitration of aromatic compounds is a highly exothermic reaction that generally uses catalysts that tend to be corrosive (e.g., HNO3 /H2SO4 ). A less corrosive employs N2 O5 as the nitrating agent as illustrated below: If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of N2O5? The reaction rate is first order in A and second order in B. Data: HRX = -370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L

Answers

Answer:

-370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L

Explanation:

The nitration of aromatic compounds is a highly exothetion that generally uses catalysts that tend to be corrosive (e.g., HNO3 /H2SO4 ). A less corrosive employs N2 O5 as the nitrating agent as illustrated below: If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of N2O5? The reaction rate is first order in A and second order in B. Data: HRX = -370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L

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rmic

Define dew point in complete sentences

Answers

The definition of dew point is Having or concerned with two points is the answer.

A student has a sample of CaSO4 hydrate and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration, and after subsequent heatings, the student finds the anhydrous compound has a constant mass of 0.3750 grams. Find the formula of the hydrate.(2 points)

Answers

Answer:

CaSO4•3H2O.

Explanation:

Let the compound be CaSO4.xH2O.

The following data were obtained from the question:

Mass of hydrated compound (CaSO4.xH2O) = 0.4813g

Mass of anhydrous compound (CaSO4) = 0.3750g

Next we shall determine the mass of the water is the hydrated compound.

This is illustrated below:

Mass of water = mass of hydrated – mass of anhydrous.

Mass of water = 0.4813 – 0.3750

Mass of water = 0.1063g

Next, we shall determine the number of mole of the anhydrous compound and the number of mole of the water present in the compound. This is illustrated below:

Molar mass of anhydrous CaSO4 = 63.5 + 32 + (16x4) = 159.5g/mol

Mass of anhydrous CaSO4 = 0.3750g

Mole of anhydrous CaSO4 =...?

Mole = mass /Molar mass

Mole of anhydrous CaSO4 = 0.3750/159.5 = 2.35×10¯³ mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O = 0.1063g

Number of mole of H2O =.?

Mole = mass /Molar mass

Mole of H2O = 0.1063/18 = 5.91×10¯³ mole

Next we shall determine the ratio of number of mole of anhydrous CaSO4 to that of H2O. This is illustrated below:

Mole of anhydrous CaSO4 = 2.35×10¯³ mole

Mole of H2O = 5.91×10¯³ mole

Ratio of anhydrous CaSO4 to H2O =>

CaSO4 : H2O => 2.35×10¯³ /5.91×10¯³

CaSO4 : H2O => 1 : 3

Therefore, for 1 mole of the anhydrous CaSO4, there are 3 moles of H2O.

Therefore, the formula for the hydrate compound CaSO4.xH2O => CaSO4•3H2O.

Suppose you have a solid that looks like gold but you believe it to be fool’s gold. The mass of the solid is 23.5 grams. When the solid is lowered into a graduated cylinder the water level rises from 47.5 to 52.2 mL. Is the simple fool’s gold

Answers

Answer:

The sample is fool's gold

Explanation:

Density is defined as the ratio between mass in grams and volume in mililiters.

A sample of pure gold has a density of 19.3g/mL.

Using Archimedes' principle, the volume of the sample is:

52.2mL - 47.5mL = 4.7mL

As the mass of the sample is 23.5g, the density is:

23.5g / 4.7mL = 5g/mL

The denisty of the sample is very different to density of pure gold, that means:

the sample is fool's gold

Arrange the following set of atoms in order of decreasing atomic size: Sn, I, Sr
Which atom has the largest atomic size?
a) Sn
b) I
c) Sr
Which atom has the smallest atomic size?
a) Sn
b) I
c) Sr

Answers

Atomic size
Sr>Sn>I

A)
B)

Write the complete electron configuration for the titanium atom.

Answers

Answer:

Explanation:

First we should fill the 4s then only 3d.

How many liters of a 1 M NaOH stock solution would you need to make 785 mL of a 215 mM NaOH dilution? (m.w. = 40.00 g/mol) Do not include units in your answer; report your answer in the requested units.

Answers

Answer:

0.1688L of the 1M NaOH stock solution

Explanation:

A 215mM = 0.215M solution of NaOH contains 0.215 moles per liter. As you want to prepare 785mL = 0.785L of the 0.215M you will need:

0.785L × ( 0.215mol / L) = 0.1688 moles of NaOH.

These moles of NaOH comes from the 1M stock solution, that means the volume of 1M NaOH solution you need is:

0.1688 moles NaOH × (1L / 1mol) =

0.1688L of the 1M NaOH stock solution

The volume of the stock solution of 1 M NaOH stock solution needed to make 785 mL of a 215 mM NaOH dilution is 0.169 L

From the question given above, the following data were obtained:

Molarity of stock (M₁) = 1 M

Volume of diluted solution (V₂) = 785 mL = 785 / 1000 = 0.785 L

Molarity of diluted solution (M₂) = 215 mM = 215 / 1000 = 0.215 M

Volume of stock solution needed (V₁) =?

The volume of the stock solution needed can be obtained as follow:

M₁V₁ = M₂V₂

1 × V₁ = 0.215 × 0.785

V₁ = 0.169 L

Therefore, the volume of the stock solution needed is 0.169 L

Learn more: https://brainly.com/question/24219233

A sample of gold weighs 1.2 oz. The sample is pounded into a thin rectangular sheet with an area of 400. sq. ft. The density of gold is 19.3 grams per cm3. What is the thickness of the foil in centimeters?Given: 28.35 g = 1 oz, 1 ft = 12 inches, 1 inch = 2.54 cm

Answers

Answer:

thickness of the gold sheet = 4.74 * 10⁻⁶ cm

Explanation:

mass of gold sample = 1.2 oz,; area of rectangular gold sheet = 400 sq. ft

Converting mass of gold from oz. to g

1 oz. = 28.35 g

mass of gold sample in grams = 1.2 * 28.35 g = 34.02 g

Converting from feet to cm;

1 feet  =  12 * 2.54 cm = 30.48 cm

1 sq. ft = (30.48)² = 929.0304 cm²

area of gold in cm² = 400 * 929.0304 cm² = 371612.16 cm²

Since the density of a solid is constant

Density = mass/volume

Volume = mass/density

where volume =area * thickness

therefore, area * thickness =mass/density

thickness = mass/(density * area)

substituting the value; thickness = 34.02 g/(19.3 gcm⁻³ *371612.16 cm²)

thickness of the gold sheet = 4.74 * 10⁻⁶ cm

What is a rate law?
A. An equation that relates the rate of a reaction to the equilibrium
position
B. An equation that relates the reaction rate to the concentrations of
the reactants
C. An equation that relates the rate constant to the temperature of a
reaction
D. An equation that relates the concentrations of reactants to the
rate constant

Answers

Answer:

B

Explanation:

The rate law is a way for us to measure the rate of a given reaction by comparing it to the concentration of the reactants. Based on the mole ratios of the reactants, we can hypothesize that adding more reactant will speed up the rate of reaction due to the simple principle that if there is more of an opportunity for molecules or atoms to collide, then the rate of reaction will be faster. Hope that helps :)

A rate law is an equation that relates the reaction rate to the concentrations of the component reactants.

Rate laws

In chemistry, rate laws are mathematical expressions that relate the rate of a reaction to the concentrations of the component reactants in the reaction.

For example, consider the following reaction:

A + B -----------> C + D

The rate law for the reaction is given as:

Rate = k[tex][A]^m[B]^n[/tex] where k = rate constant and m/n = order of reactions.

More on rate laws can be found here: https://brainly.com/question/4222261

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half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution

Answers

Answer:

2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)

Explanation:

In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.

Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-

2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)

A piece of bismuth with a mass of 22.5 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to 48.6 cm3. What is the density of indium to the proper number of significant figures

Answers

Answer:

[tex]\rho = 9.78\frac{g}{cm^3}[/tex]

Explanation:

Hello,

In this case, in order to compute the density of bismuth we need to apply the formula:

[tex]\rho =\frac{m_{Bi}}{V_{Bi}}[/tex]

Nonetheless, the volume is computed by the difference:

[tex]V_{Bi}=48.6-46.3=2.30cm^3[/tex]

Therefore:

[tex]\rho = \frac{22.5g}{2.30cm^300}\\\\\rho = 9.78\frac{g}{cm^3}[/tex]

Regards.

why homoannular diene in uv spectrometry have higher wavlenth than hetroannular diene?

Answers

Answer:

Homoanular dienes have a greater base value than heteroanular dienes

Explanation:

Woodward in 1945 gave a set of rules relating the wavelength of maximum absorption to the structure of a compound. These rules were modified by Fieser in 1959. These sets of rules describe the absorption of organic molecules in the UV region of the electromagnetic spectrum.

Each system of diene or triene has a given fixed value at which maximum absorption is expected to occur according to Woodward rules. This given fixed value is called the base or parent value. If the two double bonds are trans to each other, the diene is said to be transoid. If the two double bonds belong to different rings, the system is said to be heteroanular and the base value in each case is 215nm. If the double bonds are cis to each other (cisoid), or the two double bonds are in the same ring (homoanular), then the base value is 253nm.

Since λmax = base value + ∑ substituent contributions + ∑ other contributions, if the other contributions are not very significant, homoanular diene will have a greater λmax because of its larger base value compared to heteroanular diene. This correlates well with the fact that conjugated systems absorb at a longer wavelength.

warming oceans temperatures directly lead to all of the following except A sea level rising B coral bleaching C ocean deoxigenation D pollution

Answers

Answer:

D. pollution

Explanation:

Increase in ocean temperature is one of the major consequence of global warming which directly leads to rise in sea level,  coral bleaching and ocean deoxygenation.

Warming ocean temperatures do not leads to pollution directly whereas pollution leads to warming ocean temperatures. So, in the case of pollution, the effect is opposite.

Hence, the correct option is D.

A buffer solution is all of the following EXCEPT: a solution that contains both a weak acid and its conjugate base. a solution that resists a change in pH when a base is added. a solution that regulates pH because it is such a strong acid or base. a solution that resists a change in pH when an acid is added. All of the above are true.

Answers

Answer:

A SOLUTION THAT REGULATES pH BECAUSE IT IS SUCH A STRONG ACID OR BASE

Explanation:

A buffer solution is an aqueous solution consisting of a weak acid and its conjugate base. It is an aqueous solution used to keep the pH of solution at a nearly constant value in various chemical processes. It resists change in pH when either a strong acid or a strong base  is added. So it is very essential in various chemical applications and even in the human body as the blood pH is kept in nearly constant value by the bicarbonate buffer system in conjunction with the kidneys. The buffer solution is able to keep this nearly constant range of values because of the equilibrium between the weak acid and its conjugate base. So therefore, the incorrect statement in the options is that buffer solution is a solution that regulates pH because it is such a strong acid or base. The other options are correct.

Given Ba, Li, Na, Cs, and Be, arrange the group 1 and 2 elements in order of increasing reactivity to water (H2O).
1. Be∠Na∠ Li∠ Ba∠ Cs
2. Be∠ Li∠ Ba∠ Na∠ Cs
3. Cs∠ Na∠ Be∠ Li∠ Ba
4. Li∠ Cs∠ Na∠ Be∠ Ba

I think it's the third option, but the Beryllium and Barium are messing me up! And a brief explanation too. Thanks!

Answers

Answer:

Be∠ Li∠ Ba∠ Na∠ Cs

Explanation:

Beryllium does not react with water. It is the only alkaline earth metal that does not react with water because of its small size and high ionization energy. Beryllium differs considerably from other members of group two, its compounds when anhydrous show a considerable degree of covalent character.

As the atomic number of the group two elements increases, their ionization energies decreases and their electrode potentials become more negative hence their reactivity increases down the group. This implies that barium will have a very negative electrode potential comparable to that of the alkali metals, hence it reacts considerably with water.

The reactivity of alkali metals with water increases down the group. Lithium reacts quietly with water, sodium and potassium react with water with increasing vigour while rubidium and cesium react with water with exceptional violence.

This little explanation, is the reason behind the option chosen as the answer.

. A compound X, containing C, H, and O was found to have a relative molar mass of 6o

whilst 20.0g of X contained 8.0g of C and 1.33g of H. Calculate the empirical formula

of compound X, and hence determine its molecule formula. show calculation​

Answers

Answer:

empirical formula = CH2O

molecular formula = C2H4O2

What is the electron configuration for N (nitrogen)?
A. 1521p5
B. 1s22s23s 3p
O cly 1s22s22p3
O D. 1s 2s22p 3s
SUBMIT

Answers

Answer:

[tex]1s^22s^22p^3[/tex]

Explanation:

Nitrogen has the atomic number = 7

So, No. of electrons = 7

Electronic Configuration:

[tex]1s^22s^22p^3[/tex]

Remember that:

s sub shell holds upto 2 electrons while p sub shell upto 6

The following skeletal oxidation-reduction reaction occurs under basic conditions. Write the balanced OXIDATION half reaction.

Bi(OH)3 + NO2 → Bi + NO3-

Answers

Answer:

[tex]N^{4+}O_2+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+H_2O[/tex]

Explanation:

Hello,

In this case, for the given reaction, we first start by the writing of the oxidation states of all the involved elements:

[tex]Bi^{3+}(OH)^-+N^{4+}O^{2-}_2\rightarrow Bi^0+(N^{5+}O^{2-}_3)^-[/tex]

In such a way, we are noticing nitrogen is undergoing an increase in its oxidation state, therefore it is being the oxidized species, for which the oxidation half reaction, should be (considering basic conditions):

[tex]N^{4+}O_2+H_2O+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+2H^++2OH^-\\\\N^{4+}O_2+H_2O+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+2H_2O\\\\N^{4+}O_2+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+H_2O[/tex]

Best regards.

A 28.4 g sample of an unknown metal is heated to 39.4 °C, then is placed in a calorimeter containing 50.0 g of water. Temperature of water increases from 21.00 °C to 23.00 °C. What is the specific heat of this metal in this problem?

Answers

Answer:

THE SPECIFIC HEAT OF THE METAL IS 0.8983 J/g °C

Explanation:

In solving the problem, we have to understand that:

Heat lost by the metal = Heat gained by the water in the bomb calorimeter

First is to calculate the heat evolved from the reaction

Heat = mass * specific heat * change in temperature

Mass of water = 50 g

specific heat of water = 4.184 J/g °C

Change in temperature = 23 - 21 = 2 °C

So therefore,

Heat = 50 * 4.184 * 2

Heat = 418.4 J

Next is to solve for the specific heat of the metal;

Heat lost by the metal is the same as the heat gained by water

Heat = mass * specific heat of  metal * change in temperature

Change in temperature = 39.4 °C - 23 °C = 16.4 °C

418.4 = 28.4 * C * 16.4

C = 418.4 / 28.4 * 16.4

C = 418.4 / 465.76

C = 0.8983 J/ g °C

The specific heat of the metal is hence 0.8983 J/g °C

how many moles of a solute is present in 4.00L of an 8.30M solution​

Answers

Answer:

The number of moles of solute present in 4.00 L of an 8.30 M solution​ is 33.2

Explanation:

The Molarity (M) or Molar Concentration is the number of moles of solute per liter of solution; in other words it is the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex])  or M.

In this case:

molarity= 8.30 Mnumber of moles of solute= ?volume= 4.00 L

Replacing:

[tex]8.30 M=\frac{number of moles of solute}{4 L}[/tex]

Solving:

number of moles of solute= 8.30 M* 4 L= 8.30 [tex]\frac{moles}{liter}[/tex] * 4 L

number of moles of solute =33.2

The number of moles of solute present in 4.00 L of an 8.30 M solution​ is 33.2

Answer:

33.2 is the answer

Explanation:

did the test already :)

Which of the following best identifies where long-range order would be found?
ОООО
in amorphous solids
in crystalline solids
in thermal plasmas
in nonthermal plasmas

Answers

Answer:

in crystalline solids

Hope this answers your question, good luck

The crystalline solids represent the best identification where the long-range order should be found.

What are crystalline solids?

Crystalline solids refer to the solid where the atoms, molecules should be make the arrangement. The smallest & repeated pattern of this solid should be called as the unit cell. The unit cell should be treated as the bricks in the wall which means it should be homogenous in the nature and repeated

Therefore, The crystalline solids represent the best identification where the long-range order should be found.

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