UV spectroscopy measures the energy required to promote an electron from the ________ molecular orbital to the ________ molecular orbital.

Explanation:

1. boiling and melting point

2. electrical conductivity

3. Bond strength

4. bond length

A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

Answer:

1.07 g of water.

Explanation:

A reaction between an acid and a base makes water and a salt as product.

Our reaction is:

H₂SO₄ +  2NaOH  →  Na₂SO₄  +  2H₂O

Reactants are the acid and the base. Which is the limiting?

2.9 g . 1mol /98 g =  0.0296 moles of acid

3.53 g . 1mol / 40 g = 0.088 moles of base

2 moles of base react to 1 mol of acid

0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid

And we only have 0.0296, sulfuric acid is the limiting

Ratio is 1:2. 1 mol of acid can produce 2 moles of water.

Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.

We convert moles to mass:

0.0592 mol . 18g /mol = 1.07 g

Answer:

Around 46° F (7.8° C)

Explanation:

"Scientists have predicted that the global average temperature during the ice age was around 46 degrees Farenheit (7.8 degrees Celsius.) However, the polar regions were far colder, around 25 degrees Fahrenheit (14 degrees Celsius) colder than the global average."

Source: http://www.weforum.org

Answer:

D / 15.0 g

Explanation:

3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions

.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle

Answer:

10.77%

Explanation:

Molar mass of Cu = mass deposited/number of moles of Cu

Molar mass of Cu = 0.4391 g/6.238x10^-3 moles

Molar mass of Cu = 70.391 g/mol

%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100

%error = 10.77%

Answer:

[tex]{ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times {10}^{4} \: kelvin }}[/tex]

The temperature of the given gas is  60.95 K when it is occupying 3.70 L at 4.12 atm.

The ideal gas law can be described as a general equation of the state of an ideal gas. This equation gives the relationship between the volume and pressure of one-mole of gas equal to the multiplication of the universal gas constant and temperature.

The mathematical relationship can be shown for the ideal gas equation as:

PV = nRT

Where P is the pressure of the gas, n is the moles, V is the volume of the gas, and R is the gas constant.

Given, the volume of gas, V = 3.70 L

The  pressure of the given gas, P = 4.12 atm

The value of the gas constant, R =0.082 atmL/K mol

The number of moles of the given gas, n = 3.05 mol

Substitute the values V, R, P, and n in the ideal gas equation, and we get:

T = PV/nR

T = 4.12 × 3.70/(0.082 × 3.05)

T = 60.95 K

Therefore, the temperature of the given gas is  60.95 K.

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Answer:

A. 3.04×10^-19J

Explanation:

Hope this will help you.

Answer:

[tex]\boxed {\boxed {\sf 426 \ mL}}[/tex]

Explanation:

We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:

[tex]P_1V_1= P_2V_2[/tex]

The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.

[tex]450 \ mL * 720 \ mm \ Hg = P_2V_2[/tex]

The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.

[tex]450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2[/tex]

We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.

[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}[/tex]

[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2[/tex]

The units of millimeters of mercury (mm Hg) cancel.

[tex]\frac {450 \ mL * 720 }{760} = V_2[/tex]

[tex]\frac {324,000}{760} \ mL = V_2[/tex]

[tex]426.3157895 \ mL =V_2[/tex]

The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.

[tex]426 \ mL \approx V_2[/tex]

The volume at standard atmospheric pressure is approximately 426 milliliters.

Answer:

(B). it's metallic bonding

Answer:

There is no difference between the two.

Explanation:

They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments

Answer:

After careful consideration your answer is...

Leucippus and Democritus

*Hope I helped*

~Alanna~

Answer:

The first theories of matter were put forward by Empedocles in 450 BC, he proposed that all matter was composed of four elements - Earth, air, fire and water. Later, Leucippus and Democritus suggested matter was made up of tiny indestructible particles continuously moving in empty space.

Explanation:

Answer:

2.01

Explanation:

First, let's convert grams to moles

(Na) 22.99 + (F) 18.998 = 41.988

Every mole of NaF is  41.988 grams

21/41.988 = 0.500143 moles of NaF

For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF

0.500143/2 = 0.250071

molarity = moles/liters

0.250071/0.125 = 2.0057 M

Question Completion with Options:

O coarse...few...rapid

O fine...few...slow

O fine...multiple...rapid

O coarse...few...slow

O fine...multiple...slow

Answer:

The choice that best completes the sentence is:

O coarse...few...slow

Explanation:

Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow.  This is because of the process that starts with  recrystallization, recovery, and nucleation before growth can occur.  While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.

Answer:

A) Forms a racemic mixture of the two possible enantiomers

When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and  new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

There are three types of chemical changes:

1) inorganic changes

2)organic changes

3) biochemical changes

During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.

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 9 is the element Florine

Florine has 9 electrons as well as the 9 protons that determine its atomic number.

The ground state configuration is the lowest energy configuration.

Answer:

Mole = molecular weight / molecular mass

Mole = 16/16

Mole= 1

Answer:

[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]

The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.

Substitute these values into the formula.

[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]

[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]

Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.

[tex]molarity= 0.25 \ M \[/tex]

The molarity of the solution is 0.25 Molar and Choice D is correct.

In this equilibrium, the chemical system will shift to the right in order to produce more NH₃.

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and  equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:

[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

[tex]K = \frac{[NH_3]^2}{[H_2]^3[N_2]}[/tex]

Where [] are concentrations in equilibrium.

And Q, is:

[tex]Q = \frac{[NH_3]^2}{[H_2]^3[N_2]}[/tex]

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

Q = 1.6x10⁻⁷

As Q < K,

The chemical system will shift to the right in order to produce more NH₃

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Answer:

a) K2[Ni(CN)4]

b) Na3[Ru(NH3)2(CO3)2]

c) Pt(NH3)2Cl2

Explanation:

Coordination compounds are named in accordance with IUPAC nomenclature.

According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.

The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.

Las concentraciones en el equilibrio para la reacción química presentada son:

[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]

Consideremos la siguiente reacción química genérica:

A + B ⇄ C + D

Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.

          A + B ⇄ C + D

I          1      1      0    0

C       -x    -x     +x    +x

E      1-x    1-x    x     x

La constante de equilibrio, Kc, es:

[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]

Las concentraciones en el equilibrio son:

[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]

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Answer:

640 g

Explanation:

Step 1: Given and required data

Step 2: Calculate the moles of oxygen gas

We will use the ideal gas equation

P × V = n × R × T

n = P × V / R × T

n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol

Step 3: Calculate the mass corresponding to 20 moles of oxygen

The molar mass of oxygen is 32.00 g/mol.

20 mol × 32.00 g/mol = 640 g

Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

1.8×10⁻⁶ = x²

√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH

Answer:

by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32

The empirical formula is OsO₄ :

Explanation:

Osmium oxide contains osmium and oxygen only.

Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:

Mass of compound = 2.89 g

Mass of Os = 2.16 g

Mass of O = (Mass of compound) – (Mass of Os)

Mass of O = 2.89 – 2.16

Mass of O = 0.73 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of Os = 2.16 g

Mass of O = 0.73 g

Os = 2.16 g

O = 0.73 g

Divide by their molar mass of

Os = 2.16 / 190 = 0.011

O = 0.73 / 16 = 0.046

Divide by the smallest

Os = 0.011 / 0.011 = 1

O = 0.046 / 0.011 = 4

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Explanation:

Lanthanide series= E4

Boron=Si

Chalogen=O

Alkaline Earth metal =M9