How many atoms are in 2.32 mol2.32 mol of copper?

Answer: The ratio of AD to DB is equal to the ratio of AE to EC. In other words, the pairs of lengths are proportional.

Explanation:

Sample answer from plato

Answer:

The pairs of lengths are proportional, because, the ratios of AD and DB are the exact same, so they are equal to the ratios of the set AE and EC.

Explanation:

It is rewritten from a sample answer from Plato just to be safe from plagiarism.

Answer:

pH < 7; pH = 7; pH > 7

Explanation:

when dealing with an acidic base, you have the formula for [H3O+] > [OH-] which yields a pH < 7.

when you have a neutral base, the reactive ion concentration would be [H3O^+] = [OH^-] which yields a pH = 7.

finally, when dealing with a basic classification, the formula would be [H3O^+] < [OH^-] yields a pH > 7.

Answer:

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2]  = 1.004 M

Explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 -  2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x = [tex]\sqrt{1.6 *10^-5}[/tex] = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M

Answer:

true once light has been produce it will keep travelling jn straight parts until it hits something else

Answer:

4 m/s

Explanation:

formula is v = (KE/.5m)^1/2

there is a calculator

https://www.calculatorsoup.com/calculators/physics/kinetic.php

Explanation:

A single covalent bond can be represented by a single line between the two atoms. For instance, the diatomic hydrogen molecule, H2, can be written as H—H to indicate the single covalent bond between the two hydrogen atoms.

Answer:

pH = 5.7

Explanation:

Which is the pH of the solution?

The pH is a measurement widely used in chemistry in quality assurance of products and another analysis. Is defined as the -log [H3O+]. That means, the pH of the solution that is [H3O+] = 2x10-6 M is:

pH = -log [H3O+]

pH = -log [2x10-6 M]

Answer:

bleh

Explanation:

Answer: The correct option is A).

Explanation:

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given chemical reactions:

A): [tex]2KI+Pb(NO_3)_2\rightarrow PbI_2+2KNO_3[/tex]

The iodide of lead is generally insoluble in water. Thus, lead iodide is a precipitate.

B): [tex]2AgNO_3+Ca(CH_3COO)_2\rightarrow Ca(NO_3)_2+2CH_3COOAg[/tex]

The nitrates and acetates of all metals are soluble in water.

C): [tex]3BaCl_2+Al_2S_3\rightarrow 2AlCl_3+3BaS[/tex]

The sulfide of barium is soluble in water.

D): [tex]SrBr_2+Mg(OH)_2\rightarrow MgBr_2+Sr(OH)_2[/tex]

The hydroxide of strontium is soluble in water.

Hence, the correct option is A).

Answer and Explanation:

The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:

[tex]Kp = \frac{P^{2}_{SO_{3} } }{P_{SO_{2}} ^{2}P_{O_{2}} }[/tex]

We have the following data:

P(SO₃) = 2.6 atm

P(O₂) = 0.43 atm

We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.

Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:

[tex]P_{SO_{2} } = \sqrt{\frac{P_{SO_{3} } ^{2} }{Kp P_{O_{2} } } } = \sqrt{\frac{(2.6 atm)^{2} }{(7 x 10^{24)}(0.43 atm) } } = 1.5 x 10^{-12} atm[/tex]

Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).

Answer: trial b

Explanation:

Answer:

It is so because heat is flowing from hot body to cold body, and there is no direct contact between the body. It explains correctly the mode of transmission of thermal energy through the process of radiations.

Explanation:

Fun fact:

How does thermal energy transfer by radiation?

Radiation. All objects transfer energy to their surroundings by infrared radiation . The hotter an object is, the more infrared radiation it gives off. No particles are involved in radiation, unlike conduction.

Answer:

0.414 mole (3 sig. figs.)

Explanation:

Given grams, moles = mass/formula weight

moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)

≅ 0.414 mole (3 sig. figs.)

When we pre-weigh a glass vial to hold our sample and find its mass to be 5.010 g. Then we add our sample to the vial and reweigh it on the same balance and find that the mass has increased to 6.130 g. The mass of the sample in grams is 1.12 g.

Avogadro's number is the number of units in one mole of any substance and equals to 6.02214076 × 10²³. The units can be electrons, atoms, ions, or molecules.

No. of moles is defined as a particular no. of particles that we can calculate with the help of Avogadro’s number.

Mass of a particular product is also find out by stoichiometry of a reaction as per the no. of mole given in the reaction.

Mass is generally can be represented by units like Kg, g etc.

Given,

weigh of glass vial = 5.010 g

weigh of glass vial with sample = 6.130 g

Therefore, When we pre-weigh a glass vial to hold our sample and find its mass to be 5.010 g. Then we add our sample to the vial and reweigh it on the same balance and find that the mass has increased to 6.130 g. The mass of the sample in grams is 1.12 g.

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Answer:

H+

Explanation:

it's H+

as you see hydrogen ion it could H+

Answer:

[CO] = 0.078M

[Cl2] = 0.078M

[COCl2] = 0.477M

Explanation:

Based on the reaction:

CO(g) Cl2(g) ⇄ COCl2(g)

Where equilibrium constant, kc, is:

kc = 77.5 = [COCl2] / [CO] [Cl2]

[] represents the equilibrium concentration of each gas. The initial concentration of each gas is:

[CO] = 0.555mol/1.00L = 0.555M

[Cl2] = 0.555M

And equilibrium concentrations are:

[CO] = 0.555M - x

[Cl2] = 0.555M - x

[COCl2] = x

Where x is reaction coordinate

Replacing in kc expression:

77.5 = [x] / [0.555M - x] [0.555M - x]

77.5 = x / 0.308025 - 1.11 x + x²

23.8719 - 86.025 x + 77.5 x² = x

23.8719 - 87.025 x + 77.5 x² = 0

x = 0.477M. Right answer

x = 0.646M. False answer. Produce negative concentrations

Replacing:

And those concentrations are the equilibrium concentrations

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Being the molar mass of each compound:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

in units [tex]\frac{moles}{liter}[/tex]

then, the number of moles of HCl that react is:

[tex]6 M=\frac{number of moles of HCl}{0.01 L}[/tex]

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

[tex]mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}[/tex]

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, Mg will be the limiting reagent.

The limiting reactant in the reaction is Magnesium (Mg)

From the question,

We are to determine the limiting reactant in the reaction.

The given balanced chemical equation for the reaction is

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This means

1 mole of Mg is required to react completely with 2 moles of HCl

Now, we will determine the number of moles of each reactant present

Mass = 0.0350 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Atomic\ mass}[/tex]

Atomic mass of Mg = 24.305 g/mol

∴ Number of moles of Mg present = [tex]\frac{0.0350}{24.305}[/tex]

Number of moles of Mg present = 0.00144 mole

Concentration = 6M

Volume = 10.00 mL = 0.01 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles HCl present = 6 × 0.01

Number of moles HCl present = 0.06 mole

Since,

1 mole of Mg is required to react completely with 2 moles of HCl

Then

0.00144 mole of Mg is required to react completely with 2×0.00144 mole of HCl

2×0.00144 = 0.00288

∴ The number of moles of HCl required to react completely with the Mg is 0.00288 mole

Since the number of moles of HCl present is more than 0.00288 mole, then HCl is the excess reactant and Mg is the limiting reactant.

Hence, the limiting reactant in the reaction is Magnesium (Mg)

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Answer:

MoO₃

Explanation:

To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239,878g/mol-

11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo

Mass Mo in the oxides:

0.09830 moles Mo * (95.95g/mol) = 9.432g Mo

Mass oxygen in the new oxide:

14.151g - 9.432g = 4.719g oxygen

Moles Oxygen:

4.719g oxygen * (1mol/16g) = 0.2949 moles O

The ratio of moles of O/Mo:

0.2949molO / 0.09830mol Mo = 3

That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:

Answer:

Partial

Explanation:

A strong acid will completely ionize in water while a weak acid will only partially ionize.

Answer:

The reaction rate or rate of reaction is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time. Reaction rates can vary dramatically.

Answer:

A photon is absorbed by the atom.

More energy is emitted or absorbed for case 2

Explanation:

According to the Bohr model of the atom, electrons occur in energy levels. The energy of each level is fixed. However, electrons can absorb photons and move from a lower to higher energy level or emit photons and move from a higher to a lower energy level.

In each case, the energy absorbed or emitted is equal to the difference in energy between the two energy levels.

Since energy level 3 is much higher than energy level 2, the electron absorbs more energy in moving from energy level 1 to energy level 3 than it absorbs when moving from energy level 1 to energy level 2.

Answer:

0.80 M

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

20.05 mL of 1.000 M NaOH react.

0.02005 L × 1.000 mol/L = 0.02005 mol

Step 3: Calculate the reacting moles of HA

The molar ratio of NaOH to HA is 1:1. The reacting moles of HA is 1/1 × 0.02005 mol = 0.02005 mol.

Step 4: Calculate the concentration of HA

0.02005 moles of HA are in 25 mL.

[HA] = 0.02005 mol/0.025 L = 0.80 M

Answer:

SO < CO2 < C3H8

Explanation:

Entropy refers to the degree of disorderliness of a system. The standard molar entropy of a substance refers to the entropy of 1 mole of the substance vunder standard conditions.

The molar entropy depends on the number of microstates in the system which in turn depends on the number of atoms in the molecule.

C3H8 has 11 atoms and hence the highest number of microstates followed by CO2 having three atoms and least of all SO having only two atoms.

A. Atoms can combine in many ways to form various compounds

and molecules.

Answer:

0.186 moles

Explanation:

In order to convert grams of PCl₅ into moles, we need to use its molar mass:

Then we proceed to calculate the number of moles:

In the given question Phosphorus pentachloride is used as a catalyst for certain chemical reaction. 38.7 g of [tex]\rm PCl_5[/tex], there are 0.186 moles of [tex]\rm PCl_5[/tex].

A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy required for the reaction to occur.

To calculate the number of moles in 38.7 g of [tex]\rm PCl_5[/tex], we need to divide the given mass of [tex]\rm PCl_5[/tex] by its molar mass.

The molar mass of [tex]\rm PCl_5[/tex] can be calculated by adding the atomic masses of one phosphorus atom and five chlorine atoms:

Molar mass of  [tex]\rm PCl_5[/tex] = (1 x atomic mass of P) + (5 x atomic mass of Cl)

= (1 x 30.97 g/mol) + (5 x 35.45 g/mol)

= 208.22 g/mol

Now, we can calculate the number of moles of  [tex]\rm PCl_5[/tex]:

Number of moles = mass / molar mass

= 38.7 g / 208.22 g/mol

= 0.186 moles

Therefore, there are 0.186 moles of  [tex]\rm PCl_5[/tex] in 38.7 g of  [tex]\rm PCl_5[/tex].

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i think its 2.0

Why do i think this-If you add 0.73g to 1.3g it comes to 2.0g

Answer:

Hydrogen and fuel cell technologies power cars, buildings and more. But how ... Test your knowledge with this quiz! ... How do fuel cells generate electricity

Answer:

A

Explanation:

fuel cell is a device that converts the chemical energy from fuel into electricity via a chemical reaction with oxygen or another oxidizing agent. Batteries work in a closed system, while fuel cells require their reactants to be replenished.

Answer:

a) 0.11

b)76.9

c) 8.8

d) 1.7*10^-4

Explanation:

Step 1: Data given

K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 2: Formula of K

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K = [C]^c *[D]^d  / [A]^a * [B]^b

K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]

Step 3:

a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

1/2N2 + 3/2H2(g) ⇌ NH3(g)    =>K' =  [tex]\sqrt{K}[/tex]

K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11

b. 2NH3(g) ⇌ N2(g) + 3H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

2NH3(g) ⇌ N2(g) + 3H2(g)    =>K' = 1/K

K' = 1/(1.3*10^-2) = 76.9

c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)    

=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]

K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]

K' = 8.8

d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

2N2(g) + 6H2(g) ⇌ 4NH3(g)

K' = K²

K' = (1.3*10^-2)²

K' = 1.7 *10 ^-4

Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.

Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.

Given chemical reaction is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Equilibrium constant for this reaction is:

K = [NH₃]² / [N₂][H₂]³

K = 1.3 × 10⁻² (given)

1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)

K₁ = √K

Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

K₂ = 1/K

Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9

NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)

K₃ = 1/√K

Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8

2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)

K₄ = K²

Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴

Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.

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