alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.

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:

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: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.

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Answer:

83914.52 grams

Explanation:

Given that,

Weight of a man is 185 lb

We need to find his weight in grams

For this, we must know the relation between lb and grams.

We know that,

1 lb = 453.592 grams

To find the mass of man in grams, the step is :

185 lb = (453.592 × 185) grams

= 83914.52 grams

So, the mass of a man is 83914.52 grams.

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:

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:

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:

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:

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

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:

The number of electrons transferred in the reaction

Explanation:

Answer:

A

Explanation:

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:

1.8x10¹⁷ molecules of CO are in each breath we take

Explanation:

Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.

A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.

In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:

2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =

[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take

A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.

Now CO molecules in each breath is

[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]

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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:

3.0x10⁻²M

Explanation:

Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:

Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻

When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:

ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:

1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

1.4x10⁻⁵ = [2X]² [X]

1.4x10⁻⁵ = 4X³

3.5x10⁻⁶ = X³

0.015 = X

As [Ag⁺] is 2X:

[Ag⁺] = 0.030 = 3.0x10⁻²M

The answer is:

Answer:

H2SO4 with concentrated HCl

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.

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: 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 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

Answer:

Electrons "surround"

Explanation:

Protons and neutrons "make up" the nucleus so they are contained "within" the nucleus meaning that electrons would "surround" the nucleus as they orbit around the nucleus

Answer:

Electrons

Explanation:

Protons and nuetrons are present inside the nucleus of an atom while the electron revolve around the nucleus in different energy levels.

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:

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.

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Answer:

The rate would have doubled

Explanation:

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:

[tex]Speed = 3.30 \frac{km}{hr}[/tex]

Explanation:

Given

Distance = 1 mile

Time = 8.92 minutes

Required

Calculate Speed in km/hr

Speed is calculated as thus;

[tex]Speed = \frac{Distance}{Time}[/tex]

Substitute 1 mile for distance and 8.92 minutes for time

[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]

Convert Miles to Kilometres

If

[tex]1\ mile = 1609\ m[/tex];

Then

[tex]1\ mile = \frac{1609\ km}{1000}[/tex]

[tex]1\ mile = 1.609\ km[/tex] --- (1)

Convert minutes to hour

[tex]1\ minutes = 0.0167\ hour[/tex]

Multiply both sides by 8.92

[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]

[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)

By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have

[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]

[tex]Speed = 3.304309 \frac{km}{hr}[/tex]

[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated