Structurwl formula for (CH3)2CHCH2OH

Answer:

Koverall [NO]^2 [Br2]

Balanced chemical reaction equation;

2NO + Br2 ⇄2NOBr

Explanation:

Consider the first step in the reaction;

NO(g) + Br2(g) ⇄ NOBr2(g) fast

The second step is the slower rate determining step

NOBr2(g) + NO(g) ⇄ 2NOBr(g)

Given that k1= [NOBr2]/[NO] [Br2]

k2= [NOBr2] [NO]

The concentration of the intermediate is now;

[NOBr2]= k1[NO][Br2]

It then follows that overall rate of reaction is

Rate= k1k2[NO]^2 [Br2]

Since k1k2=Koverall

Rate= Koverall [NO]^2 [Br2]

Answer:

NOBr_2(g)+NO(g)----> 2NOBr(g)

Explanation:

Answer retrieved from ALEKS

Answer: 0.424 J/g°C

Explanation:

For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.

[tex]C=\frac{q}{m(deltaT)}[/tex]

*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.

Now that we have our equation, we can plug in our values and solve.

[tex]C=\frac{181J}{(38.8g)(36-25°C)}[/tex]

*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.

[tex]C=0.424J/g°C[/tex]

Answer and Explanation:

7.40 becomes lower or shorter than the pKa of functional groups or chains of all the 3 amino acids respectively Lys, His, as well as Asp, although since pH provided throughout the statement. However, all of the covalent chains or bonds of the three compounds or acids (amino) will serve a purpose in the pro-toned pattern.

The pH during which everyone's binding sites will mostly keep track of customer for someone's comparison is:

Lys:

[tex]pH > 10.53[/tex]

His:

[tex]pH > 6.00[/tex]

Asp:

[tex]pH > 3.65[/tex]

Answer:

Mass of solid = 189.141 gram

Explanation:

Given:

Total volume = 93 ml

Mass of liquid = 33.7 gram

Density of liquid = 0.865 g/ml

Density of solid = 3.50 g/ml

Find:

Mass of solid = ?

Computation:

Volume of liquid = Mass of liquid / Density of liquid

Volume of liquid = 33.7 / 0.865

Volume of liquid = 38.9595 ml

Volume of solid = Total volume - Volume of liquid

Volume of solid = 93 - 38.9595

Volume of solid = 54.0405 ml

Mass of solid = Volume of solid × Density of solid

Mass of solid = 54.0405 ml × 3.50 g/ml

Mass of solid = 189.141 gram

Answer:

A Reaction

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

B Non Reaction

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

Y > Q > W > Z > X

Explanation:

The first question is whether a reaction will occur base on the chemical equation below.

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

Firstly, some element are more reactive than others , base on this criteria element can be arranged  base on it reactivity .

1. Mn(s) + Ca(NO₃)₂(aq)

This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.

2. KOH(aq) +  Fe(s)

The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.

3. Pt(NO₃)₂(aq) + Cu(s)

Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.

4. Cr(s) + H₂SO₄(aq)

Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.

Base on the reaction

Q + W+ Reaction occurs

Since the reaction occurred element Q is more reactive as it displace element w from it compound.  

X + Z+ No reaction

No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.

W + Z+ Reaction occurs

Element w is more reactive than z as it displaces z form it compound.

Q+ + Y Reaction occurs

Element Y is more reactive than element  Q as it displaces Q from it compound.

Therefore, the  order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X

A. The beakers in which a chemical reaction will occur:

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

B. The beakers in which there is no reaction:

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

C. The elements from most reactive to least reactive is:

Y > Q > W > Z > X

A.

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

Elements can be arranged on the basis of reactivity:

1. Mn(s) + Ca(NO₃)₂(aq)

This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.

2. KOH(aq) +  Fe(s)

The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.

3. Pt(NO₃)₂(aq) + Cu(s)

Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.

4. Cr(s) + H₂SO₄(aq)

Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.

According to reactions:

Q + W→ Reaction occurs

Since the reaction occurred element Q is more reactive as it displace element w from it compound.  

X + Z →No reaction

No reaction occurred because element X is less reactive than Z therefore, it cannot displace z from it compound.

W + Z→ Reaction occurs

Element W is more reactive than Z as it displaces Z form it compound.

Q + Y →Reaction occurs

Element Y is more reactive than element Q as it displaces Q from it compound.

Thus, the order of reactivity from the most reactive to the least reactive will be: Y > Q > W > Z > X

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Answer:

Explanation:

CH₃C(CH₃)=CH-CH₂-CH₃ ( 2 - methylpent-2-ene)

Cl₂ ⇒ Cl⁺ + Cl⁻

Cl⁺( electrophile)

CH₃C(CH₃)=CH-CH₂-CH₃  + Cl⁺ ( electrophile)  ⇒ CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ (  intermediate )

CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ + Cl⁻  ⇒ CH₃CCl(CH₃)-C(Cl)H-CH₂-CH₃ ( final product )

It is electrophilic addition reaction.

Answer:

See exaplanation

Explanation:

For this question, we several sub-questions, lets start with the first one:

What is its molecular formula?

For this question we have to multiply the subunit by the number of subunits in the polysaccharide, so:

[tex](C_6H_1_0O_5)*2537=C_1_5_2_2_2H_2_5_3_7_0O_1_2_6_8_5[/tex]

What is its molar mass?

For this question, we have to find the molar mass of 1 subunit and then multiply by the number of subunits:

Atomic masses: C: 12 g/mol H: 1 g/mol O: 16 g/mol

Now we can multiply, the atomic masses by the number of atoms, so:

[tex](12*6)+(10*1)+(16*5)=~162~g/mol[/tex]

If we take into account the number of subunits:

[tex](162~g/mol)*2537=410994~g/mol[/tex]

What is the empirical formula of amylose?

In this case, we have to remember that the empirical formula is the smallest number of atoms. In other words, we have to simplify the formula. Therefore, the smallest formula is the subunit formula: [tex]C_6H_1_0O_5[/tex].

I hope it helps

Answer:

Option C

Explanation:

Consider the ionic equation of this chemical equation. We are given barium chloride and silver nitrate as the reactants, and silver chloride and barium nitrate as the products. We can thus conclude that the ionic equation ( not balanced yet ) should be as follows -

Ba( 2 + ) + Cl ( - ) + Ag ( + ) + NO3 ( - ) ------> AgCl + Ba( 2 + ) + NO3( - )

As you can see these compounds are present in aqueous solutions, and are thus dissociated.

______________________________________________________

Now let us take a look at the number of elements on the reactant and product sides, and balance this chemical equation out -

Ba( 2 + ) + 2Cl ( - ) + 2Ag ( + ) + 2NO3 ( - ) ------> 2AgCl + Ba( 2 + ) + 2NO3( - )

Solution = Option C!

Answer:

Explanation:

If the power of ten is positive, you move the decimal place to the right when converting  a number from scientific to standard notation

For example:

5.6894 * 10^9

10^9 is 1 with 9 zeros

5.6894 * 1000000000

5689400000.

it moved right

If the power of 10 is positive, move the decimal point to the left.

If the power of 10 is positive, move the decimal point to the right.

If the number being converted is greater than 10, move the decimal point to the left.

If the number being converted is greater than 10, move the decimal point to the right.

This is option 2

The correct answer is if the power of 10 is positive, move the decimal point to the right.

A frequently-used floating-point system in which integers are expressed as products of a number between 1 and 10 multiplied by a power of 10.

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Answer:

(i)The compression factor is 1.12

(ii) The molar volume of the gas is 2.68 L/mol

Since the compression factor is greater than 1, the attractive forces are dominating.

Explanation: Please see the attachments below

Answer:

N-Type Semiconductor

Explanation:

Answer:

See figure 1

Explanation:

On this case we have a base (methylamine) and an acid (2-methyl propanoic acid). When we have an acid and a base an acid-base reaction will take place, on this specific case we will produce an ammonium carboxylate salt.

Now the question is: ¿These compounds can react by a nucleophile acyl substitution reaction? in other words ¿These compounds can produce an amide?

Due to the nature of the compounds (base and acid), the nucleophile (methylamine) doesn't have the ability to attack the carbon of the carbonyl group due to his basicity. The methylamine will react with the acid-producing a positive charge on the nitrogen and with this charge, the methylamine loses all his nucleophilicity.

I hope it helps!

Answer: The new molarity is 2.4 M

Explanation:

According to the dilution law,

[tex]C_1V_1=C_2V_2[/tex]

where,

[tex]C_1[/tex] = concentration of pure solution = 3 M

[tex]V_1[/tex] = volume of pure solution = 200 ml

[tex]C_2[/tex] = concentration of diluted solution= ?

[tex]V_2[/tex] = volume of diluted solution= 250 ml

Putting in the values:

[tex]3M\times 200ml=M_2\times 250ml[/tex]

[tex]M_2=2.4M[/tex]

Thus the new molarity is 2.4 M

Answer:

C = 2.24x10⁻⁴ M

Explanation:

The concentration of iron in Uncle Wilbur's runoff can be calculated using Beer-Lambert law:

[tex] A = \epsilon*C*l [/tex]   (1)

Where:

A: is the absorbance of the compound

ε: is the molar absorptivity of the compound

C: is the concentration of the compound

l: is the optical path length

Since the runoff sample exhibited the same absorbance as the reference sample, we can find the concentration using equation (1):      

[tex] \epsilon*C_{1}*l_{1} = \epsilon*C_{2}*l_{2} [/tex]    (2)

Where:

Subscripst 1 and 2 refer to Uncle Wilbur's runoff and to reference sample, respectively.

l₁ = 2.41 cm

l₂ = 1.00 cm

We can find C₂ as follows:

[tex] C_{2} = \frac{C_{2i}*V_{i}}{V_{f}} [/tex]    (3)

Where:

[tex]C_{2i}[/tex]: is the initial concentration of the reference sample = 6.74x10⁻⁴ M

[tex]V_{i}[/tex]: is the initial volume = 10.0 mL

[tex]V_{f}[/tex]: is the final volume = 50.0 mL

[tex] C_{2} = \frac{6.74 \cdot 10^{-4} M*10.0 mL}{50.0 mL} = 1.35 \cdot 10^{-4} M [/tex]

Now, we can find C₁ using equation (2):

[tex] C_{1} = \frac{C_{2}*l_{2}}{l_{1}} = \frac{1.35 \cdot 10^{-4} M*1.00 cm}{2.41 cm} = 5.60 \cdot 10^{-5} M [/tex]

Finally, since the runoff solution was diluted to 100.0 mL, the initial concentration can be calculated using equation (3) for [tex]C_{1i}[/tex]:

[tex]C_{1i} = \frac{C_{1}*V_{f}}{V_{i}} = \frac{5.60 \cdot 10^{-5} M*100.0 mL}{25.0 mL} = 2.24 \cdot 10^{-4} M[/tex]

Therefore, the concentration of iron in Uncle Wilbur's runoff is 2.24x10⁻⁴ M.

I hope it helps you!

The question is incomplete, here is the complete question:

Steam reforming of methane [tex](CH_4)[/tex] produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125 L tank with 20 mol of methane gas and 10 mol of water vapor at 38 degrees celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: The equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]

Explanation:

We are given:

Initial moles of methane gas = 20 moles

Initial moles of water vapor = 10 moles

Equilibrium moles of carbon monoxide = 18 moles

Volume of the tank = 125 L

The chemical equation for the reaction of methane and water vapor follows:

                       [tex]CH_4(g)+H_2O(g)\rightleftharpoons CO(g)+3H_2(g)[/tex]

Initial:                20             10

At eqllm:          20-x         10-x            x            3x

Evaluating the value of 'x':

[tex]\Rightarrow 3x=18\\x=6[/tex]

So, equilibrium moles of methane gas = (20 - x) = [20 - 6] = 14 mol

Equilibrium moles of water vapor = (10 - x) = [10 - 6] = 4 mol

Equilibrium moles of carbon monoxide gas = x = 6 mol

The expression of equilibrium constant for the above reaction follows:

[tex]K_{eq}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]

We are given:

[tex][H_2]=\frac{18}{125}=0.144M[/tex]

[tex][CO]=\frac{6}{125}=0.048M[/tex]

[tex][CH_4]=\frac{14}{125}=0.112M[/tex]

[tex][H_2O]=\frac{4}{125}=0.032M[/tex]

Putting values in above expression, we get:

[tex]K_{eq}=\frac{(0.144)^3\times 0.048}{0.112\times 0.032}\\\\K_{eq}=3.99\times 10^{-2}[/tex]

Hence, the equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]

Answer:

Oxygen

Explanation:

In the compound tenormin, there are two highly electronegative atoms capable of accepting electrons; oxygen and nitrogen. Oxygen is more electronegative than nitrogen.

However, the oxygen atom in tenormin is bonded to carbon in a carbonyl bond. Recall that the carbonyl bond is polar and the direction of the dipole is towards the oxygen atom. Looking at the structure of tenormin, it is clear that the electron density of the bond tends towards the oxygen atom of the carbonyl group. Electron density is withdrawn from the adjacent nitrogen atom of the amine group via mesomeric and inductive mechanism towards the more electronegative oxygen atom.

On the other side of the structure, there are two oxygen atoms. These oxygen atoms are more electronegative than nitrogen thus they are more basic.

Hence the oxygen atom is the most basic atom in the compound tenormin.

Answer:

H2 is the limiting reactant.

Explanation:

From the diagram above:

H2 => White ball

O2 => Red ball

Before the reaction

H2 => White ball => 10

O2 => Red ball => 7

After the reaction

H2O => White and red ball => 10

O2 => 2

From the simple illustration above, we can see that all the H2 were used up in the reaction but there are left over of O2.

This simply means that H2 is the limiting reactant as all of it is used up in the reaction while O2 is the excess reactant as there are leftover.

Answer: the enthalpy of reaction is, -155 kJ

Explanation:-

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The final reaction is:

[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex]    [tex]\Delta H_{rxn}=?[/tex]

The intermediate balanced chemical reaction will be,

(1) [tex]CH_4(g)+O_2(g)\rightarrow CH_2O(g)+H_2O(g)[/tex]     [tex]\Delta H_1=-284kJ[/tex]

(2) [tex]CH_2O(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)[/tex]    [tex]\Delta H_2=-527kJ[/tex]

(3) [tex]H_2O(l)\rightarrow H_2O(g)[/tex]    [tex]\Delta H_3=44.0kJ[/tex]

Now multiplying (3) by 2 and adding all the equations, we get :

[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2+2\times \Delta H_3[/tex]

[tex]\Delta H_{rxn}=(-284)+(-527)+2\times (44)[/tex]

[tex]\Delta H_{rxn}=-155kJ[/tex]

Therefore, the enthalpy of reaction is, -155 kJ

Answer:

MnO4 is the oxidizing agent.

Cl is the reducing agent.

Explanation:

To know which is oxidizing agent and which is reducing agent, let us calculate the change in oxidation number of each element in the equation below:

2MnO4−(aq) + 10Cl−(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)

Please note:

1. The oxidation state of oxygen is always –2 except in peroxide where it is –1

2. The oxidation state of Hydrogen is always +1 except in hydrides where it is –1

3. The oxidation state of chlorine is always –1

For Mn:

At left hand side

MnO4 = –1

Mn + 4O = –1

O = –2

Mn + (4 x –2) = –1

Mn – 8 = –1

Collect like terms

Mn = –1 + 8

Mn = +7

At the right hand side, Mn is +2.

The oxidation state of Mn changes from +7 to +2.

For Cl:

At the left hand side

Cl = –1

At the right hand side

Cl = 0

The oxidation state of Cl changes from – 1 to 0.

The oxidation state of O and H are unchanged.

Since the oxidation state of Mn changes from +7 to +2 i.e reduced, MnO4 is the oxidizing agent.

Since the oxidation state of Cl changes from – 1 to 0 i.e increase, Cl is the reducing agent.

Answer:

[tex]1.56 \,\,mol/L[/tex]

Explanation:

Molarity of the solution measures number of moles of a solute per litre of solution.

Molarity = volume of solution in litres/number of moles of solute dissolved in solution

Volume of solution in litres = 0.86 L

Also, 1.34 mole sample of LiCl dissolves in water

So,

Molarity of the Solution = [tex]\frac{1.34}{0.86}=1.56 \,\,mol/L[/tex]

Answer:

1. Q = 8.66 KJ

2. Q = 7.58 Kcal

3. Q = 0.71 KJ

4. Q = 24.31 Kcal

Explanation:

1. The quantity of heat absorbed can be determined by:

       Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g[tex]^{0}C[/tex] and Δθ is the change in temperature.

           = 45.2 × 4.2  × (76.9 - 31.3)

           = 8656.704

∴       Q = 8.66 KJ

The quantity of heat absorbed is 8.66 KJ.

2. Q = mcΔθ + mL

Where L is the latent heat of fusion of ice = 334 J.

       = m(cΔθ + L)

       = 72.1(4.2 × 25.2 + 334)

   Q = 31712.464 J

       = 7579.466 calories

The total heat released is 7.58 Kcal.

3. Q = mcΔθ

      = 55.5 × 0.129 × (123.4 - 24.6)

     = 707.3586

The quantity of heat required to increase the temperature of gold is 0.71 KJ.

4. Q = mL

Where: L is the specific latent heat of vaporization = 533 calories.

     Q  = 45.6 × 533

      = 24304.8

The quantity of heat required to change water to steam is 24.31 Kcal.

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

Mass of oxygen required for the reaction is 16.16 g.

The equation of the reaction is;

4NH3 + 5O2 → 4NO + 6H2O

Since Number of moles = Mass/molar mass

Molar mass of ammonia = 17 g mol

Number of moles of NH3 reacted =  6.87 g/17 g mol = 0.404 moles

From the reaction equation;

4 moles of ammonia reacts with 5 moles of oxygen

0.404 moles reacts with 0.404 moles × 5 moles/4 moles  = 0.505 moles

Mass of oxygen = 0.505 moles × 32 g/mol = 16.16 g

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Answer: The actual value of [tex]\Delta G[/tex] is -618 J/mol

Explanation:

Relation of ree energy change and equilibrium constant

[tex]\Delta G=\Delta G^0+2.303\times RT\times \log K_{eq}[/tex]

where,

[tex]\Delta G[/tex] = Free energy change

[tex]\Delta G^o[/tex] = standard free energy change = +29 kJ/mol =

R = universal gas constant

T = temperature = [tex]37^0C=(37+273)K=310K[/tex]

[tex]K_{eq}[/tex] = equilibrium constant = [tex]1.02\times 10^{-5}[/tex]

[tex]\Delta G=+29000J/mol+2.303\times 8.314J/Kmol\times 310K\times \log (1.02\times 10^{-5})[/tex]

[tex]\Delta G=29000J/mol-29619J/mol=-618J/mol[/tex]

The actual value of [tex]\Delta G[/tex] is -618 J/mol

The molar mass of fructose will be "180 g/mol". To understand the calculation check below.

According to the question,

Fructose mass = 1.8 g

Water's mass = 0.100 kg

Molal freezing point depression constant, [tex]K_f[/tex] = 1.86°C/m

Freezing point change, Δ[tex]T_f[/tex] = 0.186°C

Freezing point constant, [tex]K_f[/tex] = 1.86°C/m

We know the relation,

→ Δ[tex]T_f[/tex] = i × [tex]K_f[/tex] × m

or,

         = i × [tex]K_f[/tex] × [tex]\frac{Fructose \ mass}{Fructose \ molar \ mass\times Solvent's \ mass}[/tex]

By substituting the values, we get

0.186 = 1 × 1.86 × [tex]\frac{1.8}{Fructose \ molar \ mass\times 0.100}[/tex]

By applying cross-multiplication,

Molar mass = 180 g/mol

Thus the above approach is right.          

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A 1.8 g mass of fructose is added to 0.100 kg of water and it is found that the freezing point has decreased by 0.186 °C. The molar mass of fructose is 180 g/mol.

Given:

ΔTf = 0.186 °C

Kf = 1.86 °C kg/mol

m = mass of fructose/mass of water

mass of fructose = 1.8 g

mass of water = 0.100 kg = 100 g

Use the equation:

ΔTf = Kf mi

Where:

ΔTf is the change in freezing point (in °C)

Kf is the cryoscopic constant of water (in °C kg/mol)

m is the molality of the solute (in mol/kg)

i is the van't Hoff factor

m = (mass of fructose) / (mass of water)

m = 1.8 g / 100 g

m = 0.018 mol/kg

Rearrange the equation to solve for the molar mass (M):

ΔTf = Kfmi

Substitute the values in the above equation:

0.186 = 1.86 × 0.018 × 1 / M

By applying cross multiplication,

Molar mass (M) = 180 g/mol

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Answer:

The concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰

Explanation:

Based on the different Ksp of the hydroxides of the metal ions it is possible to precipitate 1 metal as hydroxide whereas the other still in solution.

Ksp of both hydroxides is:

Mg(OH)₂ ⇄ Mg²⁺ + 2OH⁻

Ksp = [Mg²⁺] [OH⁻]² = 6.3x10⁻¹⁰

Cu(OH)₂ ⇄ Cu²⁺ + 2OH⁻

Ksp = [Cu²⁺] [OH⁻]² = 2.2x10⁻²⁰

Replacing with the concentration of the metals:

[0.27] [OH⁻]² = 6.3x10⁻¹⁰

[OH⁻] = 4.83x10⁻⁵M

That means the precipitation of Mg²⁺ begins when [OH⁻] is 4.83x10⁻⁵M

[0.7] [OH⁻]² = 2.2x10⁻²⁰

[OH⁻] = 1.77x10⁻¹⁰

As the Cu²⁺ needs a low concentration of OH⁻ to begin precipitation,

Answer:

Calcium

Explanation:

We must first put down the reaction equation as this will serve as a guide in solving the question.

M(s) + S(s) -----> MS(s)

According to the reaction equation; 1 mole of metal reacts with 1 mole of sulphur. Hence 0.09 moles of sulphur reacts with 0.09 moles of metal.

Now recall that;

Number of moles (n) = mass(m)/ molar mass(M)

Since

mass of metal reacted= 3.6g

Number of moles of metal= 0.09 moles

Then;

Making molar mass of metal the subject of the formula;

Molar mass of metal = mass of metal / number of moles of metal

Molar mass of metal = 3.6g /0.09 moles

Molar mass of metal= 40 gmol-1

The metal having a molar mass of 40gmol-1 is calcium, therefore the metal is calcium.

Answer:

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The density is  [tex]\rho = 21.1 \ g/cm^3[/tex]

Explanation:

From the question we are told that

      The lattice constant is  [tex]a = 0.387 nm = 0.387 *10^{-9} \ m[/tex]

Generally the volume of the unit cell is  [tex]V = a^3[/tex]

                                                             =>   [tex]V = [0.387 *10^{-9}]^2[/tex]

                                                                   [tex]V = 5.796 *10^{-29} \ m^3[/tex]

Converting to  [tex]cm^3[/tex]   We have  [tex]5.796 *10^{-29} * 1000000 = 5.796 *10^{-23} cm^3[/tex]

The molar mass of Tungsten is constant with a value  [tex]Z = 184 g/mol[/tex]

One mole of Tungsten contains  [tex]6.022*10^{23}[/tex] unit cells

    Where [tex]6.022*10^{23}[/tex]  is  a constant for the number of atom in one mole of a substance(Tungsten) which is known as Avogadro's constant

      Now for FCC distance  the number of atom per unit cell  is  n =  4

               Mass of Tungsten (M) =  [tex]= \frac{Z * n }{1 \ mole \ of \ Tungsten}[/tex]

=>              Mass of Tungsten (M) =  [tex]= \frac{184 * 4 }{6.023*10^{23}}[/tex]

=>              Mass of Tungsten (M) =  [tex]= 1.222*10^{-21} \ g[/tex]

 Now  

      The density of  Tungsten is  

                  [tex]\rho = \frac{M}{V}[/tex]

substituting values

                [tex]\rho = \frac{1.222*10^{-21}}{5.796*10^{-23}}[/tex]

                [tex]\rho = 21.1 \ g/cm^3[/tex]

Answer:

5.28 moles of NaOH.

Explanation:

Step 1:

Determination of the number of mole in

375.4 g of any Na2S04.

Molar mass of Na2SO4 = (2x23) + 32 + (16x4) = 142g

Mass of Na2SO4 = 375.4g

Number of mole of Na2SO4 =..?

Mole = Mass /Molar Mass

Number of mole of Na2SO4 = 375.4/142 = 2.64 moles

Step 2:

The balanced equation for the reaction.

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 3:

Determination of the number of mole of NaOH needed for the reaction.

From the balanced equation above,

2 moles of NaOH reacted to produce 1 mole of Na2SO4.

Therefore, Xmol of NaOH will react to produce 2.64 moles of Na2SO4 i.e

Xmol of NaOH = 2 x 2.64

Xmol of NaOH = 5.28 moles

Therefore, 5.28 moles of NaOH is needed for the reaction.

Answer:

Acid base titration curves shows the pH at equivalence point

Explanation:

Since the images were not shown, I will proceed to give a general description of the following acid-base titration curves:

In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H+) and hydroxide (OH-) ions will react to form water, leading to a pH of 7.

The titration curve reflects the strengths of the corresponding acid and base. If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point.

Polyprotic acids are able to donate more than one proton per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule. In the titration curve of a polyptotic acid and a strong base, The curve starts at a higher pH than a titration curve of a strong base. There is always a steep climb in pH before the first midpoint. Gradually, the pH increases until it passes the midpoint; Right before the equivalence point there is a very sharp increase in pH.

Answer:

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Explanation: