In the preparation of methyl benzoate, what is the purpose of (a) washing the organic layer with sodium bicarbonate solution

Answer:

The emission spectrum of the element is compared with known spectra.

A flame test identify an unknown element when The element's temperature of combustion is compared with other elements.

Hence, Option (D) is correct answer.

First take an unknown metal and put it under a flame. The scientist can identify the unknown substance by the color of flame they produce.  Different unknown elements produce different colors.

Example: The element Copper produce the blue flame, sodium produces a yellow flame.

Thus, from the above conclusion we can say that The element's temperature of combustion is compared with other elements.

Hence, Option (D) is correct answer.

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Answer:

Explanation:

a) Ionic

Lithium oxide

b) Covalent

[tex]$\ce{N_2O_3}$[/tex]

c) Covalent

Phosphorus trichloride

d) Ionic

[tex]Mn_2O_3[/tex]

Answer:

27.4 g/mol

Explanation:

Assuming the compound is a gas and that it behaves ideally, we can solve this problem by using the PV=nRT formula, where:

Inputting the data:

Solving for n:

With the calculated number of moles and given mass, we calculate the molar mass:

Answer:

The following properties are either physical or chemical. Which one is different from the rest based on those two categories? We chose all of the above

Answer:

sosksjsjjs

Explanation:

even i know how to type şïllily

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

Explanation:

Explanation:

The term "basic unit of matter "refers to atom

A Atom

Answer:

1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Explanation:

Let's consider the dissolving equations for the following compounds.

1. Rubidium hydroxide

RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Sodium carbonate

Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. Ammonium selenite

(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Answer:

its kcl ,option.b)

Explanation:

because its contains k+and cl- which is and ionic bond

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

[tex]2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O[/tex]

Next, we assign the oxidation numbers as follows:

[tex]2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}[/tex]

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

Answer:

a) 0.001525 mol Ca(NO₃)₂

b) 2.29 × 10²⁴ atoms Mg

General Formulas and Concepts:

Atomic Structure  

Explanation:

a)

Calcium (Ca) has a +2 charge on the periodic table. Polyatomic ion for nitrate (NO₃) has a -1 charge. We need to balance the compound:

Ca(NO₃)₂

To find the molar mass of calcium nitrate, we must pull data from the PT.

Ca has a molar mass of 40.08 g/mol

N has a molar mass of 14.01 g/mol

O has a molar mass of 16.00 g/mol

Add these all together and multiply respectively to find the molar mass of the compound:

40.08 + 2[14.01 + 3(16.00)] = 164.1 g/mol

To find the amount of moles we have given grams, divide the grams by the molar mass:

(0.2502 g) / (164.1 g/mol) = 0.001525 mol Ca(NO₃)₂

We are given 4 sig figs and our final answer is in 4 sig figs.

b)

We need to know Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc. This converts moles into atoms, molecules, formula units, etc.

We are given 3.80 moles of Mg. To find the number of atoms, we multiply Avogadro's number:

3.80 mol Mg · (6.022 × 10²³ atoms Mg/ 1 mol Mg) = 2.28836 × 10²⁴ atoms Mg

Since we are given 3 sig figs, we must round:

2.28836 × 10²⁴ atoms Mg ≈ 2.29 × 10²⁴ atoms Mg

Answer:

Solution given:

a.

we have

1 mole of calcium nitrate=164.088g

now

0.2502g of calcium nitrate=1/164.088*0.2502=

0.0015248 mole of calcium nitrate

0.2502 grams of calcium nitrate

=0.0015248moles.

b.

1 mole of magnesium=[tex]6.02214076 × 10^{23}[/tex]atoms

3.80 mole of magnesium=[tex]6.02214076 × 10^{23}×3.80=2.2884*10^{24}[/tex]atoms

[tex]\bold{\underline{2.2884*10^{24}}}[/tex]atoms

are contained in 3.80 moles of magnesium.

Answer:

∆o

Explanation:

Usually, the d level consists of a set of five degenerate d orbitals. These orbitals rema degenerate until the approach of ligands.

When ligands approach, the d orbitals are split into higher energy eg and lower energy t2g orbitals.

There exists an energy difference between this t2g and eg levels. This energy difference is called the octahedral crystal field splitting denoted as ∆o.

Answer:

Density of liquid = 1.59 g/cm³

Explanation:

We'll begin by converting 0.610 L to cm³. This can be obtained as follow:

1 L = 1000 cm³

Therefore,

0.610 L = 0.610 L × 1000 cm³ / 1 L

0.610 L = 610 cm³

Finally, we shall determine the density of the liquid. This can be obtained as follow:

Volume of liquid = 610 cm³

Mass of liquid = 972 g

Density of liquid =?

Density = mass / volume

Density of liquid = 972 / 610

Density of liquid = 1.59 g/cm³

The density of any given liquid is equivalent to the mass of the liquid divided by the volume of the liquid.

From the given information, we have:

The mass of the unknown liquid to be = 972 g

The volume of the unknown liquid to be = 0.610 L = 610 cm³

If the formula for calculating [tex]\mathbf{Density = \dfrac{Mass}{Volume}}[/tex]

Then;

[tex]\mathbf{Density \ of \ the \ unknown \ liquid = \dfrac{972 \\ g}{610 \ cm}}[/tex]

[tex]\mathbf{Density \ of \ the \ unknown \ liquid = 1.593442623 \ g/cm}[/tex]

The Density of the unknown liquid ≅ 1.593  g/cm³

In conclusion, the density of the unknown liquid is 1.593  g/cm³ to 3 significant figures.

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Answer:

E = 5.69x10⁻²⁸m

Explanation:

To solve this question we neeed to convert the wavelength in meters to energy in joules using the equation:

E = hc / λ

Where E is energy in joules, h is Planck's constant = 6.626x10⁻³⁴Js

c is light constant = 3.0x10⁸m/s

And λ is wavelength in meters = 349m

Replacing:

E = 6.626x10⁻³⁴Js*3.0x10⁸m/s / 349m

E = 5.69x10⁻²⁸m

The energy of the wavelength at which the local AM radio station is broadcasting is: A. [tex]5.69 \times 10^{-19}\;J[/tex]

Given the following data:

Scientific data:

To calculate the energy of the wavelength at which the local AM radio station is broadcasting, we would apply Einstein's equation for photon energy:

Mathematically, Einstein's equation for photon energy is given by the formula:

[tex]E=\frac{hv}{\lambda}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]E=\frac{6.626 \times 10^{-34}\times 3 \times 10^8}{3.49 \times 10^{-7}}\\\\E=\frac{1.99 \times 10^{-25}}{3.49 \times 10^{-7}} \\\\E=5.69 \times 10^{-19}\;Joules[/tex]

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Answer:

the answer is d.) potassium

Explanation:

Answer:

(d) the atoms it contains and the way these atoms are connected

Explanation:

hope it will be helpful for you

Explanation:

Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present. Physical properties can be measured without changing a substance's chemical identity.

Answer:

hydrogenphosphate I think

Answer:

56.8%

Explanation:

The reaction of the problem is 1:1. That means 1 mole of 2-methylcyclohexanone produce 1 mole of 2-methylcyclohexanol.

Percentage yield is defined as 100 times the ratio between actual yield of the reaction (5.14g) and the theoretical yield.

The theoretical yield (All reactant produce products) is obtained from the mass of the reactant as follows:

Theoretical Yield:

8.89g 2-methylcyclohexanone * (1mol/112.17g) = 0.07925 moles 2-methylcyclohexanone

Assuming all reactant produce the product in a 100% of yield, the moles of 2-methylcyclohexanol are 0.07925 moles and the mass (Theoretical yield) is:

0.07925 moles 2-methylcyclohexanol * (114.19g/mol) = 9.05g

Percentage yield:

5.14g / 9.05g * 100 = 56.8%

The percentage of the mass successfully converted into a new product is 57.2%.

The mass of the reactant (2-methylcyclohexanone) before the reduction is given as 8.89 g.

The mass of the product ( 2-methylcyclohexanol) produced is given as 5.14 g.

The percentage of the mass successfully converted into a new product is calculated as follows;

[tex]= \frac{5.14}{8.99} \times 100\% \\\\= 57.2 \ \%[/tex]

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Answer: The balanced equation for the complete oxidation reaction of ethanol is [tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

Explanation:

Combustion is the chemical process where an organic molecule reacts with oxygen gas present in the air to produce carbon dioxide and water molecules.

It is also known as an oxidation reaction because oxygen is getting added.

The chemical equation for the oxidation of ethanol follows:

[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

By stoichiometry of the reaction:

1 mole of ethanol reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 3 moles of water molecules.

Answer:

Pilot scale up

Explanation:

Preformulation studies are carried out on candidate drug molecules that show sufficient pharmacological promise in animal model(pharma approach).

It involves preliminary study of the properties of a drug which is considered a potentially active ingredient against a particular disease condition.

Scale-up is the term used to refer to the increase in the batch size of a product. This is only done after a drug has been proven successful against the target disease after extensive pilot studies.

Scale-up is the last operation carried out when a drug has passed through all stages. It is not included in preliminary preformulation studies

Explanation:

Chalk, soft, fine-grained, easily pulverized, white-to-grayish variety of limestone. Chalk is composed of the shells of such minute marine organisms as foraminifera, coccoliths, and rhabdoliths. The purest varieties contain up to 99 percent calcium carbonate in the form of the mineral calcite.

Answer:

Chalk is a salt because it has a calcium base; you can tell that it is a salt because it is powdery. It's formula is CaCO₃, or calcium carbonate.

In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.

Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase

Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.

The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids

Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

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Answer:

[tex]\boxed {\boxed {\sf 8.22 \ \textdegree C}}[/tex]

Explanation:

The question asks us to calculate the temperature of the gas at the higher volume. Since the pressure is constant, we are only concerned about volume and temperature. We will use Charles's Law. This states that the volume of a gas and the temperature of the gas have a directly proportionate relationship. The formula is:

[tex]\frac {V_1}{T_1}= \frac{V_2}{T_2}[/tex]

The gas starts at a volume of 4.17 liters and a temperature of 7.62 degrees Celsius.

[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{V_2}{T_2}[/tex]

The gas is expanded to a new volume of 4.50 liters, but the temperature is unknown.

[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{4.50 \ L}{T_2}[/tex]

We want to solve for the temperature at a higher volume. We must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

[tex]4.17 \ L * T_2 = 4.50 \ L * 7.62 \textdegree C[/tex]

The variable is being multiplied by 4.17 liters. The inverse of multiplication is division. Divide both sides by 4.17 L.

[tex]\frac {4.17 \ L * T_2 }{4.17 \ L}= \frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]

[tex]T_2=\frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]

The units of liters (L) cancel.

[tex]T_2=\frac{4.50 * 7.62 \textdegree C}{4.17 }[/tex]

[tex]T_2=\frac{34.29}{4.17 } \textdegree C[/tex]

[tex]T_2=8.22302158273 \textdegree C[/tex]

The original measurements of liters and temperature have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

The 3 to the right in the thousandth place tells us to leave the 2 in the hundredth place.

[tex]T_2 \approx 8.22 \ \textdegree C[/tex]

The temperature of the gas at the higher volume is approximately 8.22 degrees Celsius.

There are two kinds of respiratory chemoreceptors: arterial chemoreceptors, which monitor and respond to changes in the partial pressure of oxygen and carbon dioxide in the arterial blood, and central chemoreceptors in the brain. hope this helps you

have a nice day

Answer:

a) Hence, T = 207 K.

b) Hence, T2 = 226 K.

Explanation:

Now the given,

n = 0.27 moles ; P = 2.5 atm ; T = 298 K

a) γ = 5/3 since Ne is a monoatomic gas.

[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]

Hence, T = 207 K

b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]

[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]

But P = P2

[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]

This gives us:

[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]

Hence, T2 = 226 K

Answer:

Acids react with most metals to form hydrogen gas and salt. ... When an acid reacts with metal, salt and hydrogen gas are produced