Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?

Answers

Answer 1

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

Where the concentrations of the ions are the concentrations in equilibrium

For actual concentrations of a solution, you can define Q, reaction quotient, as:

Q = [Ba²⁺] [CO₃²⁻]

If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

100.0mL is the volume of the mixture of the solutions

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

A precipitate will form, BaCO₃


Related Questions

Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ

Answers

Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

Could someone help with this? Much appreciated!

Answers

Answer:

The 3rd answer down.

Na²O (sodium oxide) will be a base when exposed to water H²O

Explanation:

Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).

The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.

Consider the acid H3PO4. This acid will react with water by the following equation. H3PO4+H2O↽−−⇀H2PO−4+H3O+ What will be true of the resulting conjugate base H2PO−4? Select the correct answer below: H2PO−4 can act as an acid.

Answers

Answer:

H+/PO-4^-2

Explanation:

hydrogen has dissolved completely

In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.

What is acid - conjugate base pair?

An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.

Given chemical reaction is:

H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺

In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.

Hence, H₂PO₄⁻ is a conjugate base.

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The following reactions all have K < 1. 1) a. C6H5COO- (aq) + C6H5OH (aq) → C6H5COOH (aq) + C6H5O- (aq) b. F- (aq) + C6H5OH (aq) → C6H5O- (aq) + HF (aq) c. C6H5COOH (aq) + F- (aq) → HF (aq) + C6H5COO- (aq) Arrange the substances based on their relative acid strength.

Answers

Answer:

the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]

Explanation:

Given that :

a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]

b.  [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]

c.  [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]

Acid strength is the ability of an acid  to dissociate into a proton and an anion. Take for instance.

HA  ↔ H⁺ + A⁻

The  acid strength of the following compounds above are:

[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰

[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴

[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵

As the acid dissociation constant increases the relative acid strength also increases.

From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]

[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid

Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.

How many moles of barium sulfate are produced from 0.100 mole of barium chloride?

Answers

Answer:

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Explanation:

Barium chloride and sodium sulfate react according to the following balanced reaction:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

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

BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl : 2 moles

Then you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?

[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]

amount of moles of BaSO₄= 0.100

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of the titrant was 38.2 mL. The starting volume of the HCl solution was 20 mL. What's the concentration of the HCl? answer options: A) 0.788 M B) 0.284 M C) 3.34 M D) 0.191 M

Answers

Answer: it is A

Explanation: I am sure

Answer:

0.191 M

Explanation:

i took the test.

If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?

Answers

Molar mass:-

[tex]\\ \sf\longmapsto HC_2H_3O_2[/tex]

[tex]\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)[/tex]

[tex]\\ \sf\longmapsto 1u+24u+3u+48u[/tex]

[tex]\\ \sf\longmapsto 28u+48u[/tex]

[tex]\\ \sf\longmapsto 76u[/tex]

[tex]\\ \sf\longmapsto 76g/mol[/tex]

No of moles=0.2molGiven mass=?

[tex]\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}[/tex]

[tex]\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}[/tex]

[tex]\\ \sf\longmapsto Given\:Mass=0.2\times 76[/tex]

[tex]\\ \sf\longmapsto Given\:Mass=1.52g[/tex]

A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). Find the ratio of the rms speed of the argon molecules to that of the hydrogen. Assume hydrogen molecule has only translational degree of freedom.

Answers

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

Explanation:

The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:  

Vrms = [tex]\sqrt{3RT/M}[/tex]

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

How many atoms are in 65.0g of zinc?​

Answers

from

1moles=iatom

Mole=mass÷avogardos

Where

Avogadro's= 6.02×10²³

So moles = 65.0÷6.02×10²³

Atoms of zinc = 391.6 ×10²³

The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].

Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.

Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately  [tex]6.022 \times10^{23}[/tex]atoms per mole.

First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.

Number of moles = Mass / Molar mass

Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol

Next, multiply the number of moles by Avogadro's number to find the number of atoms.

Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]

Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol

Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.

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write the balanced nuclear equation for the radioactive decay of radium-226 to give radon-222, and determine the type of decay

Answers

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.

Ortho and para hydrogen are....... a). molecular form. b). Nuclear form. c) allotropic form. d). All​

Answers

Ortho and para hydrogen are nuclei forms

write any four characties of vertebratas?​

Answers

Answer:

1. to bend

2. to sit

3. to walk

4. To stand

Explanation:

A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell

Answers

Answer:

The answer is D) Electrolytic cell

Explanation:

An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.

When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.

These cells are the closest thing to a galvanic battery.

Answer:

b. voltaic cell

Explanation:

Founders Education answer. had to take this quiz 4 times

Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Answers

Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.

Answers

Answer:1 )T2=134°C   2) T2=339.48°C. 3)

P=817.59 mmHg.

Explanation:

1.Given ;

pressure, P1 of neon gas = 0.646 atm

temperature, T1 =242oC + 273=515oC

Volume, V1 =515ml

Volume V2= 407ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume cools at V2=407 mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(407 mL x 515 K)/515 mL= 407K.

T2= 407K -273= 134°C.   recall 0°C=273 K)

2..Given ;

pressure, P1 of neon gas = 0.633 atm

temperature, T1 =261oC + 273=534oC

Volume, V1 =694ml

Volume V2= 796ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume expands  at V2=796mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(796 mL x 534 K)/694mL= 612.48K.

T2= 612.48K -273= 339.48°C. recall 0°C=273 K

3

Given;

moles of CO2= n=0.962 mol,

temperature T=20°C=20+273 K =293 K,

volume V=21.5 L,

gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1

Using  ideal gas equation PV=nRT

P=nRT/V

P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)

P=817.59 mmHg.

Both formic acid and carbonic acid contain two hydrogen atoms. Why is the chemical formula of formic acid written HCHO₂ (with the two hydrogen atoms listed separately), while the chemical formula of carbonic acid is written H₂CO₃ (with the two hydrogen atoms written together)?

Please explain!

Answers

That is because one of the Hydrogens of Formic acid is a part of the functional group (COOH). so to make sure the reader understands that, they second H is written separately

The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Cal­culate the value of Ksp for silver carbonate from this data.

Answers

Answer:

2.3 × 10⁻⁹

Explanation:

Step 1: Write the reaction for the solution of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 2: Make an ICE chart

We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.

        CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

I                                0                 0

C                              +S               +S

E                                S                 S

The solubility product constant is:

Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹

If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?

Answers

Answer:

[tex]m=0.127g[/tex]

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]

Best regards.

how many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?​

Answers

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]

The reaction for the decomposition of ammonia is as follows:-

[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]

Calculate the mole  of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-

[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]

[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]

Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex]  are 0.2175 mol, and 0.0725 mol.

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5. Calcule las concentraciones cuando se alcanza el equilibrio si partimos de unas concentraciones iniciales [A]=[B]=1M ; [C]=[D]=0M y una constante de equilibrio de 5.

Answers

Las concentraciones en el equilibrio para la reacción química presentada son:

[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]

Consideremos la siguiente reacción química genérica:

A + B ⇄ C + D

Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.

          A + B ⇄ C + D

I          1      1      0    0

C       -x    -x     +x    +x

E      1-x    1-x    x     x

La constante de equilibrio, Kc, es:

[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]

Las concentraciones en el equilibrio son:

[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]

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What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?

Answers

Answer:

8.68

Explanation:

pOH = 8.68

all you need is contained in the sheet

Answer:

Approximately [tex]8.68[/tex].

Explanation:

The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:

[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].

On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:

[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].

At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].

Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:

[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].

Therefore, the [tex]\rm pOH[/tex] of this solution would be:

[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].

Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].

For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.

What is the molarity of an HCl solution if 25.0 mL of this solution required 17.80 mL of 0.108 M NaOH to reach the end point in a titration?

Answers

Answer:

[tex]\boxed {\boxed {\sf 0.0769 \ M}}[/tex]

Explanation:

We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:

[tex]M_AV_A= M_B V_B[/tex]

In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.

[tex]M_A * 25.0 \ mL = M_BV_B[/tex]

The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.

[tex]M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL[/tex]

We are solving for the molarity of the acid and we must isolate the variable [tex]M_A[/tex]. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

[tex]\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]

[tex]M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]

The units of milliliters cancel.

[tex]M_A= \frac{0.108 \ M * 17.80 }{25.0 }[/tex]

[tex]M_A= \frac{1.9224}{25.0 } \ M[/tex]

[tex]M_A= 0.076896 \ M[/tex]

The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.

[tex]M_A \approx 0.0769 \ M[/tex]

The molarity of the hydrochloric acid is 0.0769 Molar.

Complete the following equation of nuclear transmutation.
23892U + 126C → 24498Cf + 6 ______
Complete the following equation of nuclear transmutation.
U + C → Cf + 6 ______
A) 1n
B) 0 e
C) 0 e
D) 1H
E) 0g 0 -1 +1 1 0

Answers

Answer:

Option A. 1 0n

Explanation:

Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.

The missing part of the transmutation equation as it has been shown is 1/o n. Option A

What is nuclear transmutation?

Nuclear transmutation is the process of shifting the number of protons in an atom's nucleus to change one element into another. Nuclear processes that change one atomic nucleus into another with a different atomic number are involved.

The production of nuclear energy, radioactive decay, and the creation of new isotopes for use in science and industry all depend on nuclear transmutation, a fundamental idea in nuclear physics.

We have the equation as;

238/92 U + 12/6 C  ----> 244/98 Cf + 6 1/0 n

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In a reversible reaction, the endothermic reaction absorbs ____________ the exothermic reaction releases. A. less energy than B. None of these, endothermic reactions release energy C. the same amount of energy as D. more energy than

Answers

Answer: C. the same amount of energy as

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.

Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

[tex]A+B\rightleftharpoons C+D[/tex]

Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.

The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.

Answer:

C.) the same amount of energy as

Explanation:

I got it correct on founders edtell

Methylamine is a base because it can bond to H+. Draw Lewis structures to show how methylamine reacts with water to form a hydroxide ion.

Answers

Attached are the lewis structures,

The proton released by water bonds to the nitrogen, because it can hold charges relatively better than carbon because of it's bigger size

What 2 variables are needed to calculate the percent yield?

Answers

Answer:

isn’t it reactants and products?

Explanation:

Answer: reactants and products

Explanation:

A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?

Answers

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

Where X is reaction coordinate

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

[Cu²⁺] = 2.01x10⁻²⁶

A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for H2PO4-)

Answers

Answer:

pH of the buffer is 7.33

Explanation:

The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).

To find pH of a buffer we use H-H equation:

pH = pka + log [A⁻] / [HA]

Where A⁻ is conjugate base and HA weak acid.

For the H₂PO₄⁻ and HPO₄²⁻ buffer:

pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]

Computing values of the problem:

pH =7.21 + log [0.125M] / [0.095M]

pH = 7.33

pH of the buffer is 7.33

4NH3(g) 5O2(g)4NO(g) 6H2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 1.81 moles of NH3(g) react at standard conditions.

Answers

Answer:

-434.14 kJ

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Calculate the standard free energy change (ΔG°r) for the reaction

We will use the following expression.

ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))

ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)

ΔG°r = -959.42 kJ

Step 3: Calculate the standard free energy change for 1.81 moles of NH₃

959.42 kJ are released per 4 moles of NH₃.

[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]

What are the conditions that are favorable for extensive solid solubility of one element in another (Hume-rothery rules)

Answers

Answer:

Atomic radius less than 15%, similar structure and same valency.

Explanation:

The conditions that are favorable for extensive solid solubility of one element in another are the following.

The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.

The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.

What is Hume - Rothery rules?

Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.

Following main points are described in this rule:

Difference between the size of the solute and the solvent should be less than 15%.Electronegativity difference between the solute and solvents should be small.And they both should have same valency, means same no. of electrons in the outermost shell.

Hence size, electronegativity and valency are the conditions.

To know more about Hume - Rothery rules, visit the below link:
https://brainly.in/question/3349695

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