Please help to answer these questions please just give the option, no need explanation. Thanks

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) =

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:

1 × V₁ = 0.215 × 0.785

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

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

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

The third energy level will hold 18​ electrons. Hence, option (D) is correct.

The electrons surrounding an atom are located in regions around the nucleus called “energy levels”.

An energy level represents the 3-dimensional space surrounding the nucleus where electrons are most likely to be.

The general formula is that the nth shell can hold up to 2(n²) electrons in principle.

Therefore, The third energy level will hold 18​ electrons. Hence, option (D) is correct.

Learn more about Energy level here ;

https://brainly.com/question/17396431

#SPJ2

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!

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.

Answer:

Explanation:

(1) 1-methylbutane

(2) 1,1,3-trimethylhexane

(3) 5octyne

(4)2-ethyl-1-propanol

(5)2,2-dimethyl1-3-butanol

Answer:

n= 0.08186

{He}2s^2 2p^6

Explanation:

PV=nRT

n=PV/RT

n= (1.220 atm)(4.3410 L) / (0.0821 atm*L/mol*K)(788.0 K)

n=0.08186

As for the electron configuration:

Ne:

{He} 2s^2 2p^6

or long hang:

1s^2 2s^2 2p^6

[tex][Ne]=1s^22s^22p^{10}[/tex]

Given:

To find:

1.

The pressure of the gaseous boron trifluoride = P = 1.220 atm

The volume of the gas in bulb = V = 4.3410 L

The moles of the  gaseous boron trifluoride = n

The temperature of gaseous boron trifluoride = T = 788.0 K

Using an ideal gas equation:

[tex]PV = nRT\\\\1.220 atm\times 4.3410 L=n\times 0.0821 atm L/mol K\times 788.0 K\\\\n=\frac{1.220 atm\times 4.3410 L}{0.0821 atm L/mol K\times 788.0 K}\\\\n=0.08186 mol[/tex]

The moles of gaseous boron trifluoride is 0.08186 moles.

2.

The atomic number of neon atom = 10

The electronic configuration in the ground state is the most stable arrangement of the electrons in the lowest energy levels.

The ground state electronic configuration of neon is:

[tex][Ne]=1s^22s^22p^{10}[/tex]

Learn more about the ideal gas equation and electronic configuration here:

brainly.com/question/1056445?referrer=searchResults

brainly.com/question/5524513?referrer=searchResults

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:

Answer:a

Explanation:

The three characteristics of minerals are that they are solid, have definite crystal structure and definite chemical composition.

Minerals are defined as a chemical compound  which has a well -defined composition and possesses a specific crystal structure.It occurs naturally in the pure form.

If a compound occurs naturally in different crystal structure then each structure is considered as a different mineral.The chemical composition of a mineral varies  depending on the presence of small impurities which are present in small quantities.

Some minerals  can have variable proportions of two or more chemical elements  which occupy equivalent position in the crystal structure.It may also have variable composition which is split into separate species.

Physical properties of minerals include color,streak, luster,specific gravity  and cleavage.

Learn more about minerals ,here:

https://brainly.com/question/1333886

#SPJ6

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.

Answer:

[tex]Q=2582J=2.58kJ[/tex]

Explanation:

Hello,

In this case, for us to compute the absorbed heat, we apply the following equation:

[tex]Q=m_{Al}Cp_{Al}(T_2-T_1)[/tex]

Whereas we use the mass, specific heat and temperature change for the piece of aluminium, thus, we obtain:

[tex]Q=63.1g*0.930\frac{J}{g*\°C}*(67.0-23.0)\°C\\ \\Q=2582J=2.58kJ[/tex]

It is positive as the heat is entering, therefore the temperature raises.

Best regards.

Answer:

Water isn't combustible. There is nothing you can add to it other than gasoline that will make it even remotely combustible. Now, through electrolysis, it can be broken down into hydrogen and oxygen, which could be burned in an internal combustion engine.

because  the water  in itself does not produce energy

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:

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 :)

Answer: products; reactants

Explanation: as the endothermic reactions are tye one which absorbs energy

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

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.

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.

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)

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.

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.

Learn more about solid here: https://brainly.com/question/17431076

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

Hold on, our servers are swamped. Wait for your answer to fully load.

rmic

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.

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

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.

Answer:

empirical formula = CH2O

molecular formula = C2H4O2

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.

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.