Using the periodic table to locate each element, write the electron configuration of(b) Cu;

Answer:

2g/ml

Explanation:

d=m/v

d=20/10

d=2

Answer:

\(\boxed {\boxed {\sf N_2}}\)

Explanation:

Let's examine this reaction, especially the coefficients.

\(N_2+3H_2 \rightarrow 2NH_3\)

First of all, remember that reactants are used to make the product. They are found on the left of the arrow usually. Therefore, N₂ and H₂ are the reactants, while NH₃ is the product. We can automatically eliminate NH₃ from our answer choices, because it is simply not a reactant.

Next, look at the coefficients.

Therefore, for the reaction to work, there must be 1 mole of N₂ and 3 moles of H₂.

We have 3.2 moles of N₂ and 5.4 moles of H₂.

Divide each amount given by the required amount for completion.

Therefore, there is enough nitrogen to complete the reaction 3.2 times, but only enough hydrogen for 1.8 times. If everything is completely reacted, we will run out of hydrogen and have excess nitrogen.

The Answer is Solid

Explanation:

the particles in a solid are vibrating very fast and are packed together so they're independent one each other.

Answer:

Water is needed for iron to rust.....

Explanation:

As you can see the liquid in tube A is water liquid in tube B is a drying agent and the liquid in Tube C is Boiled water with oil above .

The only change was in Tube A, a red-brown deposit which is rust.

And Tube B and C didn't have any change.

Which results that only water can change iron to rust.

U 2 can help me by marking as brainliest.........

Answer:

1. D 2.. A 3. C 4. B

Explanation:

just took the quiz

We can classify each rock as intrusive or extrusive and basaltic or granitic as follows; Rock A - Extrusive, Basaltic, Rock B - Intrusive, Granitic, Rock C - Extrusive, Basaltic, and Rock D - Extrusive, Basaltic.

Rock A - Extrusive, Basaltic

Rock A is described as having dark-colored large grains, which indicates a fine-grained texture, typical of extrusive rocks. The presence of dark-colored large grains suggests a basaltic composition.

Rock B - Intrusive, Granitic

Rock B is described as having large crystals and a high percentage of silica, which suggests a coarse-grained texture, typical of intrusive rocks. The high silica content and large crystals indicate a granitic composition.

Rock C - Extrusive, Basaltic

Rock C is described as having a fine-grained texture and being light-colored, which are characteristic features of extrusive rocks. The fine-grained texture suggests a rapid cooling, typical of extrusive rocks, and the light color is indicative of a basaltic composition.

Rock D - Extrusive, Basaltic

Rock D is described as being from the Hawaiian volcano area and having no visible crystals. This indicates a glassy texture, typical of rapidly cooled extrusive rocks. The origin from a volcanic area and glassy texture suggest a basaltic composition.

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The pH of a solution is a measure of its acidity or basicity. It is determined by the concentration of hydrogen ions (H+) in the solution. The pH scale ranges from 0 to 14, with 0 being highly acidic, 7 being neutral, and 14 being highly basic.

In this scenario, the chemist is titrating a solution of aniline with a solution of a strong base. Aniline is a weak base, so when it reacts with the strong base, it will undergo neutralization. The balanced chemical equation for the reaction is:
Aniline + Base → Salt + Water
To calculate the pH of the base solution after adding a certain volume of the base solution, we need to consider the initial concentration of the aniline solution and the volume of the base solution added.

Here are the steps to calculate the pH:
1. Convert the volume of the base solution added to liters, if necessary.
2. Determine the initial number of moles of aniline in the solution using the initial concentration and volume of the aniline solution.
3. Use stoichiometry to determine the number of moles of base reacted with the aniline.
4. Calculate the final concentration of the base solution by dividing the moles of base reacted by the total volume of the solution (aniline solution + base solution).
5. Calculate the pOH of the base solution using the formula: pOH = -log[OH-] (where [OH-] is the concentration of hydroxide ions).
6. Subtract the pOH from 14 to obtain the pH of the base solution.

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(a) The % solids in the leach feed is 90%.

(b) The wt.% Ni in the leach residue is 0%.

(a) The % solids in the leach feed:

To calculate the % solids in the leach feed, we need to consider the mass balance of the process.

Given:

Ore feed rate: 5,000 TPD

Ni extraction: 90%

Leach solution production rate: 6,500 TPD

We can start by calculating the amount of Ni entering the leach solution:

Ni entering leach solution = Ore feed rate * Ni content

= 5,000 TPD * 1.5 wt.% = 75 TPD

Since the Ni extraction is 90%, the Ni content in the leach solution after extraction can be calculated as:

Ni in leach solution = Ni entering leach solution * Ni extraction

= 75 TPD * 90% = 67.5 TPD

Next, we need to calculate the amount of solids in the leach feed. We are given that the solids weight decreases by 10% during the leach. Let's assume the initial solids weight in the leach feed is S TPD.

After the leach, the solids weight becomes 90% of the initial weight, i.e., 0.9S TPD.

Now, we can set up a mass balance equation for the Ni in the leach feed:

Ni in leach feed = Ni in leach solution + Ni in leach residue

Since we know the Ni in the leach solution (67.5 TPD) and the Ni content in the leach feed (1.5 wt.%), we can solve for the solids weight (S):

Ni in leach feed = S TPD * 1.5 wt.%

S = Ni in leach feed / (1.5 wt.%)

= 67.5 TPD / (1.5 wt.%)

= 4,500 TPD

Finally, we can calculate the % solids in the leach feed:

% solids in leach feed = (S TPD / Ore feed rate) * 100

= (4,500 TPD / 5,000 TPD) * 100

= 90%

Therefore, the % solids in the leach feed is 90%.

(b) The wt.% Ni in the leach residue:

To calculate the wt.% Ni in the leach residue, we can use the information from part (a) and the mass balance equation:

Ni in leach residue = Ni in leach feed - Ni in leach solution

= 4,500 TPD * 1.5 wt.% - 67.5 TPD

= 6,750 TPD - 67.5 TPD

= 6,682.5 TPD

The weight of the leach residue can be calculated by subtracting the weight of the leach solution from the weight of the leach feed:

Weight of leach residue = Ore feed rate - Leach solution production rate

= 5,000 TPD - 6,500 TPD

= -1,500 TPD (negative value indicates there is no residue)

Since the weight of the leach residue is negative, it means there is no leach residue produced. Therefore, the wt.% Ni in the leach residue is 0%.

(a) The % solids in the leach feed is 90%.

(b) The wt.% Ni in the leach residue is 0%.

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Answer:

nu

Explanation:

It has five naturally occurring stable isotopes. The most common stable isotope of zinc is Zn-64. Zn-64 has such a long half-life that it radioactive properties are virtually non-existant. ... Zinc is a hard metal, but becomes very malleable above 100 degrees Celsius.

Answer:

4.70 grams of NH4Cl is needed to prepare 350 mL of a 0.25 M ammonium chloride solution.

We need approximately 4.68 grams of NH4Cl to prepare a 0.25 M ammonium chloride solution with a volume of 350 mL.

To determine the mass of NH4Cl needed to prepare the solution, we us use the formula:

m=M x V x MM ... (i)

where,

m= mass in grams

M=molarity of solution

MM= molar mass of compound

V= volume in litres

The number of moles of NH4Cl needed can be calculated using:

  Moles = Molarity x Volume ...(ii)

  Moles = 0.25 mol/L x 0.350 L

  Moles = 0.0875 mol

Hence we can replace M x V with number of moles in equation i.

The molar mass of NH4Cl is :

  Molar mass of NH4Cl = (1 x 14.01 g/mol) + (4 x 1.01 g/mol) + (1 x 35.45 g/mol)

  Molar mass of NH4Cl = 53.49 g/mol

We have all the variables

Putting them in equation i.

Hence,

  Mass (g) = Moles x Molar mass

  Mass (g) = 0.0875 mol x 53.49 g/mol

  Mass (g) = 4.68 g

Therefore, you would need approximately 4.68 grams of NH4Cl to prepare a 0.25 M ammonium chloride solution with a volume of 350 mL.

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When a moderate amount of a weak acid is placed in aqueous solution  the HA which is a nonionized acid considered as the greatest concentration. Option (A) is correct.

A weak acid is defied as an acid which is partially dissociates into its ions in an aqueous solution or in the water. It is evident that the conjugate base of a weak acid is a weak base while the conjugate acid of a weak base is considered as a weak acid. When a weak acid is placed in aqueous solution a non ionized form is in greatest concentration as the weak acid dissociates smaller amount. These are the acids which don't completely dissociate in the solution. A weak acid generally not considered as the strong acid as it is not dissociates completely.

The chemical reaction can be written as,

HA (aq.)  + \(H_{2} O\) (l)   ------------>  \(A^{-}\) (aq.)  +  \(H_{3}O ^{+}\) (aq.)

In this chemical reaction,  HA is considered as the greatest concentration.

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The complete question is,

when a moderate amount of a weak acid is placed in aqueous solution, which of the following will be present in the greatest concentration?

select the correct answer below:

A. the nonionized acid

B. the hydronium ion

C. the conjugate base of the acid

D.  impossible to predict

The reaction for which the enthalpy change under standard conditions corresponds to a standard enthalpy of formation is

option d. CO(g) + H2O(g) → CO2(g) + H2(g).

Enthalpy is a thermodynamic property of a system that represents the sum of its internal energy and the product of its pressure and volume, often used to describe heat transfer in chemical reactions.

Standard enthalpy is the enthalpy change that occurs when a reaction takes place under standard conditions, which are defined as a temperature of 298 K (25°C), a pressure of 1 bar, and a concentration of 1 mol/L.

The reaction for which the enthalpy change under standard conditions corresponds to a standard enthalpy of formation is

option d. CO(g) + H2O(g) → CO2(g) + H2(g).

This is because the reaction involves the formation of one mole of CO2(g) and one mole of H2(g) from one mole of CO(g) and one mole of H2O(g) under standard conditions. The enthalpy change for this reaction is equal to the standard enthalpy of formation of CO2(g) and H2(g) minus the standard enthalpy of formation of CO(g) and H2O(g). Therefore, it corresponds to a standard enthalpy of formation.

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Answer:

The answer is A Tuberculosis

The capacity of stationary phases used in high performance liquid chromatography (HPLC) to distinguish between polar column and nonpolar chemicals can be used to classify them.

Extremely polar chemicals, which are difficult to separate under reversed-phase conditions, can be retained and separated successfully using hypercarb columns. This application has shown that: Polar chemical retention is good in hypercarb columns. The techniques created on Hypercarb columns are solid.

(We define a bond to be polar if there is a difference between the electronegativity of the atoms in the bond that is more than 0.4. The bond is effectively nonpolar if the difference in electronegativity is smaller than 0.4.) The molecule is nonpolar if there are no polar bonds.

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Answer:

Option B

Explanation:

While adding salts to water is a pretty safe chemical reaction, in its elemental state, each component can be highly reactive. Sodium and chlorine both react violently with water, but are stable when their ions are together in a salt.

The activation energy of the reaction is approximately 71.46 kJ/mol, rounded to 2 decimal places.

To find the activation energy (Ea) of the reaction, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and activation energy. The Arrhenius equation is given by:

k = A * e^(-Ea/RT)

Where:

k is the rate constant

A is the pre-exponential factor (frequency factor)

Ea is the activation energy

R is the gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin

We have two sets of data:

At 600 K, k1 = 3.4 m^(-1)s^(-1)

At 750 K, k2 = 31.0 m^(-1)s^(-1)

Taking the natural logarithm (ln) of both sides of the Arrhenius equation, we can rearrange the equation to solve for the activation energy:

ln(k) = ln(A) - (Ea/RT)

We can create two equations using the given data:

ln(k1) = ln(A) - (Ea/(R * 600))

ln(k2) = ln(A) - (Ea/(R * 750))

Subtracting the second equation from the first eliminates the ln(A) term:

ln(k1) - ln(k2) = (Ea/R) * ((1/600) - (1/750))

Simplifying further:

ln(k1/k2) = (Ea/R) * ((750 - 600) / (600 * 750))

Now we can solve for Ea:

Ea = (R * (ln(k1/k2))) / ((750 - 600) / (600 * 750))

Using the given values and the appropriate units:

Ea = (8.314 J/(mol·K) * ln(3.4/31.0)) / ((750 - 600) / (600 * 750))

Converting the units from J to kJ:

Ea = (8.314 × 10^(-3) kJ/(mol·K) * ln(3.4/31.0)) / ((750 - 600) / (600 * 750))

Ea ≈ 71.46 kJ/mol

Therefore, the activation energy of the reaction is approximately 71.46 kJ/mol, rounded to 2 decimal places.

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There are 13 atoms in the product

Given

Reaction

H2SO4 + 2KOH --> K2SO4 + 2H2O

Required

The number of atoms

Solution

In a chemical equation, there are reactants on the left and products on the right

Reactants : H2SO4 + 2KOH

Products : K2SO4 + 2H2O

The number of atoms is determined by their reaction coefficient and the subscript of the atoms in the compound

K2SO4 (coefficient = 1) :

K = 2 atoms

S = 1 atom

O = 4 atoms

Total atoms = 7 atoms

2H2O(coefficient = 2) :

H = 2 x 2 = 4 atoms

O = 2 x 1 = 2 atoms

Total atoms = 6 atoms

Total = 13 atoms

Answer:

Barium permanganate | Ba(MnO4)2 - PubChem.

Description: Barium permanganate appears as ...

Synonyms: BARIUM PERMANGANATE7787-3...

Molecular Formula: Ba(MnO4)2 or BaMn2O8

Molecular Weight: 375.2 g/mol

Explanation:

An acidic solution can be defined as one that has an excess of H+ ions .So the correct option is option B.

Acidity is a property of a substance that describes its ability to donate hydrogen ions (H+). A substance with a high concentration of H+ ions is considered acidic. In aqueous solutions, the concentration of H+ ions is balanced by the concentration of hydroxide ions (OH-). When the concentration of H+ ions is greater than the concentration of OH- ions, the solution is acidic.

Option A is incorrect because an excess of OH- ions in a solution makes it basic or alkaline, not acidic.

Option C is incorrect because the anion is not directly related to acidity. An anion is a negatively charged ion that is formed when an atom gains one or more electrons.

Option D is incorrect because electrochemistry deals with the transfer of electric charge in chemical reactions. Acidity is a broader concept that involves the concentration of H+ ions in a solution.

Option E is incorrect because the loss of electrons yielding a positively charged ion is called oxidation, which is not directly related to acidity.

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Answer:

Due to the accumulation of static charges/due to static electricity

2.0 mol C₆H₁₂O₆

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

Answer:

2200d cc

Explanation:

Almost all the commonly using cleaning products in homes and offices are nothing but they are the disinfectants. The chemical products that destroy all bacteria, fungi, and viruses on surfaces are known as Disinfectants.

A disinfectant is defined as the antimicrobial agent which is usually applied on the surface of some objects in order to destroy the microorganisms residing on it. Chlorine bleach is the most powerful disinfectant.

Any substance which acts on non-living objects to kill germs, like viruses, bacteria, etc. is called the disinfectant.

Thus the correct option is B.

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Explanation:

The reaction between magnesium ribbon and dilute hydrochloric acid is a classic example of a single replacement reaction, which can be represented by the following chemical equation:

Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

In this reaction, magnesium (Mg) reacts with hydrochloric acid (HCl) to form magnesium chloride (MgCl2) and hydrogen gas (H2).

The rate of this reaction can be altered by changing the concentration of the hydrochloric acid. This is because the rate of a chemical reaction depends on the concentration of the reactants. Specifically, the rate of the reaction is proportional to the concentration of the reactants raised to some power, which is determined by the reaction's rate law.

In this reaction, the rate law can be expressed as:

Rate = k [Mg] [HCl]^x

Where k is the rate constant and x is the order of the reaction with respect to hydrochloric acid. The order of the reaction with respect to magnesium is one, since the concentration of magnesium does not change during the reaction.

When the concentration of hydrochloric acid is increased, the rate of the reaction increases because there are more hydrochloric acid molecules available to collide with magnesium atoms and react. This means that the value of x is greater than zero and the reaction is dependent on the concentration of hydrochloric acid.

Conversely, when the concentration of hydrochloric acid is decreased, the rate of the reaction decreases because there are fewer hydrochloric acid molecules available to react with magnesium. This means that the value of x is less than one and the reaction is not entirely dependent on the concentration of hydrochloric acid.

Answer:

\(\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}\)

Answer:

m = 16.67 g of Hg

Explanation:

Let's write the overall decomposition reaction that is taking place:

2HgO ---------> 2Hg + O₂

According to the balanced reaction, we can see that the mole ratio between HgO and Hg is the same (2:2 or 1:1), therefore, we can assume that the moles reactants of HgO would be the moles produced of Hg. So, in order to get the grams of mercury, we need to determine the moles first.

To get the moles, we need to use the following expression:

moles = mass / MM   (1)

The molar mass of HgO can be calculated using the atomic weights, which are:

Hg: 200.59 g/mol ; O: 15.999 g/mol

MM HgO = 200.59 + 15.999 = 216.589 g/mol

The moles are:

moles HgO = 18 / 216.589 = 0.0831 moles

As we stated before, moles reactants are the same moles produced, so:

moles HgO = moles Hg = 0.0831 moles of Hg.

Finally, to get the mass, we just solve the mass from (1):

m = moles * MM   (2)

m = 0.0831 * 200.59

Hope this helps

The composition and growth method, which enable the development of BCN films predominating between insulating BN and semimetallic graphite, determine the properties of BCN thin films. The influence of synthesis settings on various BCN characteristics are investigated in this regard.

Van der Waals bonds, of which hydrogen bonding is an example, are crucial to the characteristics of water and to how it behaves in biology. Dipole-dipole interactions with bonds between hydrogen and other atoms are thought to be the cause.

Both inorganic materials like water and biological compounds like DNA and proteins include hydrogen bonding. Van der Waals attractions, which depend on tiny variations in the electron concentrations, can happen between any two or more molecules.

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The formula unit and name for the compounds formed when each pair of ions interacts are as follows:

1. Al³⁺ and OH⁻:

  - Formula unit: Al(OH)₃

  - Name: Aluminum hydroxide

2. Mg²⁺ and SO₄²⁻:

  - Formula unit: MgSO₄

  - Name: Magnesium sulfate

3. Li⁺ and NO₃⁻:

  - Formula unit: LiNO₃

  - Name: Lithium nitrate

4. NH₄⁺ and Cl⁻:

  - Formula unit: NH₄Cl

  - Name: Ammonium chloride

1. Al³⁺ and OH⁻:

The formula unit for the compound formed when Al³⁺ and OH⁻ ions interact is Al(OH)₃. The name of this compound is aluminum hydroxide. It consists of one aluminum ion (Al³⁺) and three hydroxide ions (OH⁻) to achieve charge balance.

2. Mg²⁺ and SO₄²⁻:

The formula unit for the compound formed when Mg²⁺ and SO₄²⁻ ions interact is MgSO₄. The name of this compound is magnesium sulfate. It consists of one magnesium ion (Mg²⁺) and one sulfate ion (SO₄²⁻).

3. Li⁺ and NO₃⁻:

The formula unit for the compound formed when Li⁺ and NO₃⁻ ions interact is LiNO₃. The name of this compound is lithium nitrate. It consists of one lithium-ion (Li⁺) and one nitrate ion (NO₃⁻).

4. NH₄⁺ and Cl⁻:

The formula unit for the compound formed when NH₄⁺ and Cl⁻ ions interact is NH₄Cl. The name of this compound is ammonium chloride. It consists of one ammonium ion (NH₄⁺) and one chloride ion (Cl⁻).

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Answer:

c. Gas particles don't interact

Explanation:

Ideal gas particles do not interact with each other.

The gas molecules do not interact with each other except for colliding with each other. Gases expand to completely fill a container; they would not if they were attracted to each other. That being said, they obviously still do, as a result of Dalton's Law.

answer:asap is the best seller asap is the more complicated i tems