What is the volume of HCl (r= 1.17 g / cm ^ 3, omega = 36 \%) required to precipitate silver in the form of AgCl from 2, 000 g of an alloy containing 22% by weight of Ag, if used 1.5 times the amount of precipitating reagent?​

Answer:

2

Explanation:

i did this

Answer:

generally occurs on Ser, Thr, and/or Tyr side chains and to a lesser extent on the His side chain

Kevin can effectively deliver an update by sending a detailed EMAIL to Jill

The question is incomplete, the complete question is shown in the image attached to this answer

Answer:

2,3,3-trimethylhexane

Explanation:

IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.

According to this system of nomenclature;

A comma is used to separate two numbers.

A hyphen is used to separate a number from a letter.

Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.

Answer:

λ = 150 nm

Explanation:

For C-O bond rupture:

The required energy to rupture C-O bond = bond energy of C-O bond

= 799 kJ/mol

[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]

[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]

[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]

Recall that the wavelength associated with energy and frequency is expressed as:

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

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

[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]

[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]

λ = 150 nm

the rate would decrease because the reactants are being depleted.

Answer:

a. 7.24m

b. 5.15M

c. 53.4mL of the solution would contain this amount of ethanol.

Explanation:

Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.

Molarity, M, are the moles of solute per kg of solvent

To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:

a. Moles ethanol -Molar mass: 46.07g/mol-:

25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol

kg solvent:

100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg

Molality:

0.54265 moles ethanol / 0.075kg = 7.24m

b. Liters solution:

100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L

Molarity:

0.54265 moles ethanol / 0.1053L = 5.15M

c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =

53.4mL of the solution would contain this amount of ethanol

Answer:

111.95mL of HNO3 are needed to prepare the buffer

Explanation:

We can solve this equation using H-H equation for bases:

pOH = pKb + log [HA+] / [A]

Where pOH is the pOH of the solution

pOH = 14 - pH = 14 - 8.970 = 5.03

pKb is the pKb of NH3 = 4.74

[HA+] could be taken as moles of NH4+

[A] as moles of NH3

The NH3 reacts with nitric acid, HNO3, as follows:

NH3 + HNO3 → NH4+ + NO3-

That means the moles of HNO3 added = X = Moles of NH4+ produced

And moles of NH3 are initial moles NH3 - X

Initial moles of NH3 are:

0.125L * (0.374mol/L) = 0.04675 moles NH3

Replacing in H-H equation:

pOH = pKb + log [HA+] / [A]

5.03 = 4.74 + log [X] / [0.04675-X]

0.29 = log [X] / [0.04675-X]

1.95 =  [X] / [0.04675-X]

0.0912 - 1.95X = X

0.0912 = 2.95X

X = 0.0309 moles

We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:

0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed

In mL:

111.95mL of HNO3 are needed to prepare the buffer

Answer:

A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.

Explanation:

You have to put energy into a molecule to break its chemical bonds. The amount needed is called the bond energy. After all, molecules don't spontaneously break

Answer:

c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.

Answer:

A. 3.04×10^-19J

Explanation:

Hope this will help you.

Answer:

26.4g

Explanation:

The balanced chemical equation as stated in this question is given as follows:

Fe2O3 + 3CO → 2Fe + 3CO2

According to this balanced equation, 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).

We need to change 50.26 g of ferric oxide to moles by using the formula;

mole = mass/molar mass

Molar mass of Fe2O3 = 56(2) + 16(3)

= 112 + 48

= 160g/mol

mole = 50.26/160

mole = 0.314mol of Fe2O3

If 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).

Hence, 0.314 mol of Fe2O3 will completely react with (0.314 × 3) mol of CO

0.314 × 3 = 0.94 mol of CO

molar mass of CO = 12 + 16 = 28g/mol

mole = mass/molar mass

mass = mole × M.M

mass = 0.94 × 28

mass = 26.4g of CO

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]

[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]

[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]

Hence, the percent abundance of O-18 is:  

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]  

[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]

[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]                                                              

Hence, the percent abundance of oxygen-18 is:

[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

Answer:

kcl and pb(ch3coo)2

The precipitate is PbCl2

Explanation:

Let us take the options provided one after the other;

In the first case, we have;

LiBr(aq) + NH4NO3(aq) ----> LiNO3(aq) + NH4Br(aq)

You can see that no precipitate is formed here.

In the second case;

2KCl(aq) + Pb(CH3COO)2(aq) ----> PbCl2(s) + 2CH3COOK(aq)

The precipitate here is PbCl2.

Answer:

C) as ionic strength increases, the value of the activity coefficient increases.

Explanation:

The effective concentration of ions available for reactions is known as the activity of the ion.

The activity coefficient important in chemistry because it accounts for the deviation of a solution from ideal behaviour.

The activity of a chemical species is defines as the product of concentration and activity coefficient.

Following the Debye–Hückel limiting law; log γ = −0.509z2I1/2. The ionic strength of a solution tends to increase as the activity coefficient (γ) of the ion decreases.

Answer:

and I'll call you when the party's over

quiet when I'm come in home

when I'm all alone

Answer:

Explanation:

Don't you know too much already?

I'll only hurt you if you let me

Call me friend but keep me closer (call me back)

And I'll call you when the party's over

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= [tex]\frac{100.0 \ g}{114.33\ g/mole}[/tex]

= [tex]0.8747 \ moles[/tex]

We know that 16 moles of [tex]CO_2[/tex] formation associates with -11018 kJ of heat, then

0.8747 moles [tex]CO_2[/tex] formation associates with,

= [tex]-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat[/tex]

= [tex]-0.0547\times 11018[/tex]

= [tex]-602.69 \ KJ \ heat[/tex]

Answer: True

Explanation:

Phosphagens are high energy storage compounds that are usually found in the tissue of animals.

Based on the question, the molecules have similar functions in different organisms such as the fact that they can accept phosphoryl groups from ATP in a situation where the ATP is in excess. Also, they donate phosphoryl groups to ADP in order for the regeneration of ATP.

Answer:

25 g

Explanation:

Step 1: Write the balanced equation

HNO₃ + NaHCO₃ ⇒ NaNO₃ + H₂O + CO₂

Step 2: Calculate the reacting moles of HNO₃

100.0 mL of 3.0 mol/L HNO₃ reacted.

0.1000 L × 3.0 mol/L = 0.30 mol

Step 3: Calculate the reacting moles of NaHCO₃

The molar ratio of HNO₃ to NaHCO₃ is 1:1. The reacting moles of NaHCO₃ are 1/1 × 0.30 mol = 0.30 mol.

Step 4: Calculate the mass corresponding to 0.30 moles of NaHCO₃

The molar mass of NaHCO₃ is 84.01 g/mol.

0.30 mol × 84.01 g/mol = 25 g

Answer:

18.82 mg

Explanation:

From the given information:

The molar mass of CO2  is calculated as follow

= (12 + (16 ×2))

= 44

The mass of carbon is determined by dividing the mass no of carbon from co2 by the molar mass of CO2,  followed by multiplying it by 69.00 mg

= [tex](\dfrac{12}{44}\times 69 )[/tex]

=(0.2727 × 69 )

= 18.82 mg

Answer:

hlo.......................,

Answer:

Solution given:

1 mole of KCl[tex]\rightarrow [/tex]22.4l

1 mole of KCl[tex]\rightarrow [/tex]74.55g

we have

0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g

74.55g of KCl[tex]\rightarrow [/tex]22.4l

10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

1 mol of any gas contains 22.4L of volume at STP

Volume of KCl:-

Answer:

Both Tech A and Tech B.

Explanation:

Catalyst is an element used to start chemical reaction but is not used in the reaction. Catalysts material used in catalytic converter include Rhodium, Palladium and platinum. The pre cat and post cat oxygen sensors helps determine converter efficiency.

Answer:

Explanation:

B

[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]

Balanced Equation:-

[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]

Answer: Promotion of electrons is accompanied by a release of energy because of absorption of photon.

Explanation:

Promotion of electrons occurs when an electron accepts or absorbs a photon which leads to it's movement from a lower energy level orbital to a higher energy orbital.

According to Bohr, the electrons was restricted to certain energy levels and was thought to move along certain circular orbits around the nucleus. These energy levels were identified by means of principal quantum number, n. The wave mechanics model of atom does not restrict the electrons to a certain energy levels only. Instead it describes a region around the Nucleus called orbitals where there is a possibility of finding an electron with a certain amount of ENERGY.

The energy levels are composed of one or more orbitals and the distribution of electrons around the nucleus is determined by the number and kind of energy levels that are occupied.

Bohr made an assumption that an electron emits energy in the form of radiation when it moves from a higher to a lower permitted orbit, this produces a line in the atomic emission spectrum. Since the energies of the higher and lower orbits are fixed, the line will be of a particular energy and frequency.

Answer:

one mole of water (6.022 x 10 23 molecules) has a mass of 18.02 g. One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.

Explanation:

• The mole (or mol) represents a certain number of objects.

• SI def.: the amount of a substance that contains the same

number of entities as there are atoms in 12 g of carbon-12.

• Exactly 12 g of carbon-12 contains 6.022 x 10 23 atoms.

• One mole of H 2O molecules

contains 6.022 x 10 23 molecules.

• 1 mole contains 6.022 x 10 23 entities (Avogadro’s number)

• One mole of NaCl contains 6.022 x 10 23 NaCl formula units.

• Use the mole quantity to count formulas by weighing them.

• Mass of a mole of particles = mass of 1 particle x 6.022 x 1023

Mass of 1 H atom: 1.008 amu x 1.661 x10-24 g/amu = 1.674 x10-24 g

Mass of 1 mole of H atoms:

1.674 x10-24g/H atom x 6.022 x1023H atoms = 1.008 g  

• The mass of an atom in amu is numerically the same

as the mass of one mole of atoms of the element in grams.

• One atom of sulfur has a mass of 32.07 amu;

one mole of S atoms has a mass of 32.07 g

Answer:

10.2 mg

Explanation:

Step 1: Calculate the total amount of water she drank

1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.

If she drank 1.4 L of water per day, the total amount of water she drank is:

730 day × 1.4 L/day = 1022 L

Step 2: Calculate the amount of Pb in 1022 L of water

The concentration of Pb is 10 ppb (10 μg/L).

1022 L × 10 μg/L = 10220 μg

Step 3: Convert 10220 μg to milligrams

We will use the conversion factor 1 mg = 1000 μg.

10220 μg × 1 mg/1000 μg = 10.2 mg