Aqueous ammonia is added to a mixture of silver chloride and water. Given that Kf for the reaction between Ag+ and NH3 is large, which of the following are true?
A) The free ions are favored over the complex ion.
B) The complex ion is favored over solid silver chloride.
C) The free Ag+ ion is unstable.
D) More silver chloride will precipitate.

Answer:

4.16x10⁻³m

Explanation:

Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.

As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.

Moles glucose:

There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.

Molar mass glucose:

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

72.06 + 12.10 + 96 = 180.16g/mol

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Molality of the solution:

4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =

The molarity of the solution is 4.16x10⁻³m

We know that the molarity refers to the ratio that arise between the moles of a solute.

Since there are 100 g of solvent so here the kg should be 0.1.

Likewise there is 75 mg so it should be 0.075g

Now the Molar mass glucose should be

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

So,

= 72.06 + 12.10 + 96

= 180.16g/mol

Now

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Now finally

Molality of the solution:

= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent

=4.16x10⁻³m

learn more about molarity here: https://brainly.com/question/13011592

Answer:

A    ⇒ 20

B    ⇒ 40

C    ⇒ Ca

D    ⇒ 10

E     ⇒ 9

F     ⇒ F

Explanation:

edge 2021

Answer:

the person above is correct

Explanation:

Answer:

The correct answer is 206 ml.

Explanation:

Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,  

M₁V₁ = M₂V₂

Now putting the values in the formula we get,  

0.0887 × 69.6 = 0.0224 M × V₂

V₂ = 0.0887 × 69.6 / 0.0224

V₂ = 275.6 ml

Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,  

= 275.6 ml - 69.6 ml = 206 ml

The question is incomplete, the complete question is;

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75

Answer:

A nucleus with a A nucleus with a neutron:proton ratio of 1.49

A nucleus with a mass of 187 and an atomic number of 75

Explanation:

The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1

Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.

Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.

Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.

Answer:

-A nucleus with a neutron:proton ratio of 1.49

-The nucleus of Sb-123

-A nucleus with a mass of 187 and an atomic number of 75

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.

Answer:

Yes, same piece can be used.

Explanation:

The same piece can be used for two different atoms are acceptable because both atoms has 7 electrons in their outermost valance shell. Both atoms belong to same group i. e. halogens so same piece can be used for both atoms. If the atoms belong to different groups and they have different number of electrons in their outermost shell so using same piece will be a problem so it is recommended to use different pieces for different atoms.

The use of the same modeling piece for chlorine and fluorine has been accepted as it has consisted of the same properties and belongs to the same group.

Chlorine and fluorine have been the elements of group 17. The elements are halogens with the presence of 7 valence electrons.

The elements have been belonging to the same group and have the same number of valence electrons thus resembling each other in the chemical properties.

Since both the elements are similar to each other, the use of the same piece for two different atoms has been acceptable.

For more information about the modeling piece, refer to the link:

https://brainly.com/question/701369

Answer:

12-crown-4

Explanation:

We must recall that any structural moiety in organic chemistry having the R-O-R unit is an ether. If the oxygen form a ring in which they are sandwiched in between carbon atoms, the compound is known as a crown ether. The name emanates from the close resemblance of the compound to an actual crown.

If we want to name the crown ether, we first count the number of carbon atoms present and the number of oxygen atoms present. The correct name is now, total number of carbon + oxygen atoms -crown- number of oxygen atoms, in this case; 12-crown-4, hence the answer.

Answer:

12-crown-4

Explanation:

Answer:

H2SO4 with concentrated HCl

Answer:2.62 L

Explanation:

A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.

An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.

By using ideal gas equation,

P₁ V ₁ ÷ T = P₂V₂ ÷ T

1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50

V₂ = 1 × 2.62 × 50 ÷ 25 × 2

V₂ = 2.62 liters.

Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

To learn more about ideal gas law follow the link below;

https://brainly.com/question/6534096

#SPJ5

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

Answer:

2, hydrogen

Explanation:

i think

Answer:

Answer is not hydrogen

Explanation:

did the test and got it wrong

Answer:

There are fifteen molecular orbitals in benzene filled with electrons.

Explanation:

Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;

There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.

There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons

The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.

All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.

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:

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]

Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.

Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:

    0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

Answer:

V= n/M

Explanation:

From;

π = nRT/V = MRT

Where;

n= number of moles

R= gas constant

T= absolute temperature

M= molar mass

V= volume of the solution

π= osmotic pressure

Thus;

nRT/V = MRT

nRT = VMRT

V= nRT/MRT

V= n/M

Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]

Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:

[tex]\Delta H=\Delta E+\Delta(PV)[/tex]

So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.

Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.

Analysing each reaction above:

Reaction A has the same number of moles in reagents and products, so  enthalpy change and internal energy change will be equal;

Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.

Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.

Answer:

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

Answer:

Explanation:

A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.

Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is

2-hydroxy-2,3-dimethylbutane

     H  OH   H    H

      |     |       |      |

H - C - C -   C  - C - H

      |     |       |      |

     H  CH₃  CH₃ H

From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain

Explanation:

After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the  last step when balancing a redox reaction under basic condition.

The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

the chemical equation of ethanol as fuel is

C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)

the preparation process of ethanol from cassava is

cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol

hope this helps u

pls mark as brainliest .-.

Answer:

14.297 g

Explanation:

From the question;

1 mo of the compound requires 1320.0 kJ

From the molar mass;

1 ml of the compound weighs 30.55g

How many grams requires 617.30kJ?

1 ml = 1320

x mol = 617.30

x = 617.30 / 1320

x = 0.468 mol

But 1 mol = 30.55

0.468 mol = x

x = 14.297 g

Answer:

b. 0°C and 1 bar

Explanation:

Hello,

In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.

Best regards.

Answer:

B.

Explanation:

A chemical reaction that has the general formula of nA → (A)n is best classified as a polymerization reaction.

Answer:

B. Polymerization

Explanation:

I'm just smart

Answer:

3.11

Explanation:

Any buffer system can be described with the reaction:

[tex]HA~->~H^+~+~A^-[/tex]

Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]

With all this in mind, we can write the reaction for our buffer system:

-) Nitrous acid: [tex]HNO_2[/tex]

-) Sodium nitrate: [tex]NaNO_2[/tex]

[tex]HNO_2~->~H^+~+~NO_2^-[/tex]

In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).

We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:

-) moles of [tex]HNO_2[/tex]:

[tex]mol=0.150~M*0.045~L=0.00657[/tex]

-) moles of [tex]NO_2^-[/tex]:

[tex]mol=0.175~M*0.020~L=0.0035[/tex]

The total volume would be:

0.020 L + 0.045 L = 0.065 L

With this in mind, we can calculate the molarity of each compound:

-) Concentration of [tex]HNO_2[/tex]

[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]

-) Concentration of [tex]NO_2^-[/tex]

[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]

The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:

[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]

The final pH value would be 3.11

I hope it helps!

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.

Nitrous acid is a weak acid and nitrate ion is its conjugate base.

It is a solution used to resist abrupt changes in pH when acids or bases are added.

We do so by multiplying the molar concentration by the volume in liters.

HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol

NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol

The total volume will be the sum of the volumes of each solution.

V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L

HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M

NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M

We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.

pH = pKa + log [NaNO₂]/[HNO₂]

pH = 3.16 + log 0.0538/0.104 = 2.87

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

Learn more about buffers here: https://brainly.com/question/24188850

Answer:

The rate would have doubled

Explanation:

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to: