A chemist dissolves 14.0 g of calcium hydroxide in one beaker of water, and 17.0 g of iron(III) chloride
in a second beaker of water. Everything dissolves.
When the two solutions are poured together, solid iron(III) hydroxide precipitates.
1. Write a balanced molecular equation.
2. Determine the identity of the limiting reactant.
3. Predict the mass of iron(III) hydroxide product.

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:

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.

Electrons are a type of subatomic particle with a negative charge. Protons are a type of subatomic particle with a positive charge. ... Neutrons are a type of subatomic particle with no charge (they are neutral). Like protons, neutrons are bound into the atom's nucleus as a result of the strong nuclear force.

Answer:

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

Answer:

Quantization is the process of constraining an input from a continuous or otherwise large set of values (such as the real numbers) to a discrete set (such as the integers).

Explanation:

Quantization refers to the situation where an electromagnetic field consists of discrete energy parcels, photons.

In Chemistry , the concept that a system cannot have any possible energy value but instead is limited to certain specific energy values (states). This states depend on the specific system in question.

Under this system, Energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy).

Hence, Quantization refers to the situation where an electromagnetic field consists of discrete energy parcels, photons.

Learn more about Quantum here ;

https://brainly.com/question/16746749

#SPJ2

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:

A

Explanation:

The iron corrodes so it oxidized

Answer:

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

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S[tex]_{Generation[/tex] = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S[tex]_{Generation[/tex] = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S[tex]_{Generation[/tex] = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S[tex]_{Generation[/tex] = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

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:

Ethyn gas ( acetylene gas )

Explanation:

All group II carbides react with water to form ethyn gas apart from beryllium which produces methane gas.

[tex]{ \sf{CaC_{2(s)} + 2H _{2} O_{(l)} → Ca(OH) _{2(s)} + C _{2} H _{2(g)} }}[/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.

the rate would decrease because the reactants are being depleted.

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:

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

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:

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:

The molar mass is "185.44 g/mol".

Explanation:

According to the question,

The moles of NaOH will be:

= [tex]\frac{0.315}{1000}\times 25.72[/tex]

= [tex]8.1018\times 10^{-3} \ moles[/tex]

Number of moles of an acid will be:

= [tex]\frac{8.1018\times 10^{-3}}{3}[/tex]

= [tex]2.7006\times 10^{-3} \ moles[/tex]

We know that,

⇒ [tex]Moles = \frac{Mass}{Molar \ mass}[/tex]

hence,

Molar mass of unknown acid will be:

= [tex]\frac{Mass \ in \ g}{Moles}[/tex]

= [tex]\frac{0.5008}{2.7006\times 10^{-3}}[/tex]

= [tex]185.44 \ g/mol[/tex]

[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:

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

Answer:

A. 3.04×10^-19J

Explanation:

Hope this will help you.

Answer:

Help you with what hmm I do not know what you are talking about

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

Answer:

A reaction is the time that is required for a chemical reaction to go essential to completion

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:

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:

White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively

Explanation:

I hope it will help you

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:

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