During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. Draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization.

-----OCH3

Answer:

3.6487g

CuCl2 moles reacted = (0.15×250)/1000

according to balanced chemical equation

precipitated Cu(OH)2 moles = Reacted CuCl2 moles

molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5

mass of precipitate = (97.5 × 0.15×250)/1000

= 3.648g

Answer:

C₂H₄O₂

Explanation:

Step 1: Divide each percentage by the atomic mass of the element

C: 40.00/12.01 = 3.331

H: 6.67/1.01 = 6.60

O: 53.33/16.00 = 3.333

Step 2: Divide all the numbers by the smallest one

C: 3.331/3.331 = 1

H: 6.60/3.331 ≈ 2

O: 3.333/3.331 ≈ 1

The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.

CH₂O × 2 = C₂H₄O₂

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

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Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

Answer:

a)  [tex]V_1=5ul[/tex]

b)  [tex]v=20ul[/tex]

Explanation:

From the question we are told that:

initial Concentration [tex]C_1=50mg/ml[/tex]

Final Concentration [tex]C_2=10mg/ml[/tex]

Final volume needs [tex]V_2 =25ul[/tex]

Generally the equation for Volume is mathematically given by

[tex]C_1V_1=C_2V_2[/tex]

[tex]V_1=\frac{C_1V_1}{C_2}[/tex]

[tex]V_1=\frac{10*25}{50}[/tex]

[tex]V_1=5ul[/tex]

Therefore

The volume of buffer needed is

[tex]v=V_2-V_1\\\\v=25-5[/tex]

[tex]v=20ul[/tex]

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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Answer:

1087.84 J

Explanation:

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

Mass of metal (Mₘ) = 70 g

Temperature of metal (Tₘ) = 80 °C

Mass of water (Mᵥᵥ) = 100 g

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) = 24.6 °C

Heat lost by metal (Qₘ) =?

NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Thus, we shall determine the heat gained by water. This can be obtained as follow:

Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

Qᵥᵥ = 100 × 4.184 (24.6 – 22)

Qᵥᵥ = 418.4 × 2.6

Qᵥᵥ = 1087.84 J

Thus, the heat gained by water is 1087.84 J.

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Qᵥᵥ = 1087.84 J

Qₘ = 1087.84 J

Therefore, the heat lost by the metal is 1087.84 J

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.

Let's use the following expression to calculate the heat absorbed by the water.

Qw = c × m × ΔT

Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ

where,

According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.

Qw + Qm = 0

Qm = -Qw = -10.8 kJ

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

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Answer:

Low value for copper recovery

Explanation:

The percentage recovery is obtained from;

Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100

Note that the fact that some of the copper nitrate solution splashed out of the beaker means that some amount copper has been lost from the system. This loss of copper leads to a lower value of copper recovered from solution.

Answer:

NaoH= sodium hydroxide

Answer:

[tex]\lambda=3451*10^{10}m[/tex]

Explanation:

From the question we are told that:

Energy state [tex]e=3.50 eV[/tex]

Time [tex]t=2ms[/tex]

Generally the equation for energy of Photon is mathematically given by

[tex]E=e-e_0[/tex]

[tex]E=3.6*10^{-19}J[/tex]

[tex]E=5.7*10^{-19}J[/tex]

Generally the equation for Wave-length of Photon is mathematically given by

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

[tex]\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}[/tex]

[tex]\lambda=3451*10^{10}m[/tex]

The clumping of magnesium sulfate means that the wrong kind of drying agent have been used for the sample.

A drying agent is also referred to as a desiccant. It is a substance that is used to remove moisture from a sample. We must recall that the drying agent to be used must not react with the sample.

Since the magnesium sulfate was found to clump together at the bottom of the flask, it means that the wrong kind of drying agent have been used for the sample.

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Answer:

81°C

111°C

cyclohexane

Explanation:

Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.

Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.

Answer:

Cm = n/V

n(H2SO4) = 245/98 = 2.5 mol

Cm(H2SO4) = 2.5/1 = 2.5 M

Explanation:

Answer:

False

Explanation:

The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE

This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺

Answer:

Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

Answer:

B

Explanation:

Ammonia is considered a base as it's pH is 11

Answer from Gauthmath

The  compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.

Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.

Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.

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Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

Moles HCl added:

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

Moles NaOH to titrate the excess:

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

Moles of acid that were neutralized:

0.0105 moles - 0.0010 moles =

1. homogenous: sugar solution

2. heterogeneous: sand solution

3. compound: water

4. physical change: ice melting

5. element: hydrogen

6. chemical change: burning fire

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

Explanation:

So, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

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Answer:

atom &bond

Explanation:

atom is strong due to the bond

Answer:

8.65 atm

Explanation:

Using ideal law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (Latm/molK)

T = temperature (K)

According to the information given in this question;

V = 18.0 L

n = 6.20 moles

R = 0.0821 Latm/molK

T = 33°C = 33 + 273 = 306K

P = ?

Using PV = nRT

P × 18 = 6.20 × 0.0821 × 306

18P = 155.76

P = 155.76/18

P = 8.65 atm

Answer:

5 × 10⁵ molecules (500,000 molecules)

Explanation:

Step 1: Convert 3 × 10⁻¹⁶ g to moles

We will use the molar mass of TCDD (321.97 g/mol).

3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol

Step 2: Convert 9 × 10⁻¹⁹ mol to molecules

The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).

9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules

Answer:

Have you ever touched a fish? Most fish will feel a bit rough - due to their scales. Some, like sharks, will feel like sandpaper. Even fish with small, smoother scales will feel a bit like that. Amphibians don’t have scales, and most species will be wet to some degree - they have to keep their skin moist or they’ll die. A few groups, like toads and newts, have rougher skin, which is heavier and thicker, which allows them to retain moisture better away from water.

Functionally, the big thing about amphibian skin is that it is semi-permeable. Amphibians can breathe through their skin - all amphibians can get some oxygen through their skin, but some species of salamanders get all their oxygen that way - they have no lungs or gills. The skin can also allow water in - sort of like a paper towel. The bad thing is that other chemicals can pass through the skin, too - pollutants and other chemicals tend to affect amphibians far more than they do other groups.

Amphibians also shed their skin - fish do not. People don’t tend to see frogs shedding their skin often, though, since they eat it to regain nutrients and other resources in the skin.

Finally, since amphibian skin offers no defense against predators in the way that scales do, and limited barrier against disease the way non-amphibian skin does (shedding helps), the skin of many amphibians contain toxins, and some of them have anti-fungal properties (typically due to symbiotic bacteria). Many species have evolved chemical defenses in the skin, while others have glands that produce toxins that can be secreted outside of the skin.

The skin can withstand dessication more than the fish.

They have moist skin used as respiratory surface during deep sleep / hibernation.

They have moist skin due to secretion of mucus by glands under the skin.

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Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

Answer:

Explanation:

I am almost sure that the products are identical.