9. How many moles of H2SO4 are there in 98 g of H2so4

Answer:

C7H5F3

Explanation:

The following data were obtained from the question:

Mass of Carbon (C) = 57.53g

Mass of Hydrogen (H) = 3.45g

Mass of Fluorine (F) = 39.01g

The empirical formula of the compound can be obtained as follow:

C = 57.53g

H = 3.45g

F= 39.01g

Divide each by their molar mass

C = 57.53/12 = 4.79

H = 3.45/1 = 3.45

F = 39.01/19 = 2.05

Divide each by the smallest

C = 4.79/2.05 = 2.3

H = 3.45/2.05 = 1.7

F = 2.05/2.05 = 1

Multiply through by 3 to express in whole number

C = 2.3 x 3 = 7

H = 1.7 x 3 = 5

F = 1 x 3 = 3

Therefore, the empirical formula for the compound is C7H5F3

Answer:

B.) 2x

Explanation:

Hello,

In this case, we can apply the following rule of three, knowing that 0.1 dm³ equals x and 0.2 dm³ is the unknown:

[tex]0.1dm^3\longrightarrow x\\0.2dm^3 \longrightarrow ?[/tex]

Thus, solving for the unknown we find:

[tex]?=\frac{0.2dm^3*x}{0.1dm^3} \\\\?=2*x[/tex]

Therefore, the answer is B.) 2x.

Best regards.

Answer:

- N2 does not exist as a liquid at pressures below 0.127 atm.

- N2 is a solid at 16.7 atm and 56.5 K.

- N2 is a liquid at 1.00 atm and 73.9 K

- N2 is a gas at 0.127 atm and 84.0 K.

Explanation:

Hello,

At first, we organize the information:

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Best regards.

Hey there!

For this we can use the combined gas law:

[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]

We are only working with pressure and temperature so we can remove volume.

[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]

Solve for T₂.

[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]

[tex]2atm * T_{2}={2.2atm}*27C[/tex]

[tex]T_{2}={2.2atm}\div2atm*27C[/tex]

[tex]T_{2}=1.1*27C[/tex]

[tex]T_{2}=29.7C[/tex]

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

Answer:

2.29x10⁻¹² is Ksp of the salt

Explanation:

The Ksp of the metal hydroxide is:

M(OH)₂(s) ⇄ M²⁺ + 2OH⁻

Ksp = [M²⁺] [OH⁻]²

As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:

pOH = 14- pH

pOH = 14 - 10.22

pOH = 3.78

pOH = -log[OH⁻]

1.66x10⁻⁴ = [OH⁻]

And [M²⁺] is the half of [OH⁻], [M²⁺] = 8.30x10⁻⁵

Replacing in Ksp formula:

Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²

Ksp = 2.29x10⁻¹² is Ksp of the salt

Answer:

1.41 M

Explanation:

First we must use the information provided to determine the concentration of the aluminum hydroxide.

Mass of aluminum hydroxide= 320mg = 0.32 g

Molar mass of aluminum hydroxide= 78 g/mol

Volume of the solution= 5.00 ml

From;

m/M= CV

Where;

m= mass of aluminum hydroxide= 0.32 g

M= molar mass of aluminum hydroxide = 78 g/mol

C= concentration of aluminum hydroxide solution = the unknown

V= volume of aluminum hydroxide solution = 5.0 ml

0.32 g/78 g/mol = C × 5/1000

C = 4.1×10^-3/5×10^-3

C= 0.82 M

Reaction equation;

Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)

Concentration of base CB= 0.82 M

Volume of base VB= 1.60 ml

Concentration of acid CA= the unknown

Volume of acid VA= 2.80 ml

Number of moles of acid NA = 3

Number of moles of base NB= 1

Using;

CA VA/CB VB = NA/NB

CAVANB = CBVBNA

CA= CB VB NA/VA NB

CA= 0.82 × 1.60 × 3/ 2.80 ×1

CA= 1.41 M

Therefore the concentration of HCl is 1.41 M

Answer:

1661μL of a 6M HCl you need to add

Explanation:

pH is defined as -log[H⁺] ([H⁺] =10^{-pH}), the initial and final concentrations of [H⁺] you need are:

Initial [H⁺] = 10^{-3.5} = 3.16x10⁻⁴M H⁺

Final [H⁺] = 10^{-1} = 0.1M H⁺

In moles, knowing volume of the solution is 0.1L:

Initial [H⁺] = 0.1L ₓ (3.16x10⁻⁴mol H⁺ / L) = 3.16x10⁻⁵moles H⁺

Final [H⁺] = 0.1L ₓ (0.1mol H⁺ / L) = 0.01 moles H⁺.

That means, moles of H⁺ you need to add to the solution is:

0.01mol - 3.16x10⁻⁵moles = 9.9684x10⁻³ moles of H⁺.

A solution of HCl dissociates in H⁺ and Cl⁻ ions, that means moles of HCl added are equal to moles of H⁺. As you need to add 9.9684x10⁻³ moles of H⁺ = 9.9684x10⁻³ moles of HCl:

9.9684x10⁻³ moles of HCl ₓ (1L / 6mol) = 1.6614x10⁻³L

In μL:

1.661x10⁻³L × (1x10⁶μL / 1L) =

Answer:

3.7136g/cm³

Explanation:

Density is defined as the ratio between mass of a substance and its volume.

First, we will find the mass of the piece of the metal that is the difference between mass of metal + flask and mass of empty flask. That is:

Mass metal:

143.557g - 44.232g = 99.325g of the metal

Now, to find its volume you must know first the volume of the flask that can be obtained from the mass of water in the filled flask, that is:

153.617g - 44.232g = 109.385g of water = cm³ of water

In the second experiment, the mass of water = its volume is:

226.196g - 143.557g = 82.639g = 82.639cm³ of water

That means the volume the piece of metal is occupying is:

109.385cm³ - 82.639cm³ = 26.746cm³ of piece of metal

And its density is:

99.325g / 26.746cm³ =

Answer:

The high concentration of the iron ion ensures that the reaction is favored to go to the right

Explanation:

According to Le Chateliers principle, when a constraint such as a change in concentration, pressure or volume is imposed on a chemical system in equilibrium, the system will readjust itself in order to annul the constraint. This is done by shifting the position of equilibrium, either to the left hand side or the right hand side depending on the requirement of the system based on the imposed constraint. Equilibrium his now reestablished in the system as a result of this readjustment.

Since there is a high(excess) concentration of iron in the system, it can be safely assumed that the equilibrium concentration of the FeNCS2+ is equal to the initial concentration of SCN- ion present. This implies that the equilibrium position moves towards the right hand side according to Le Chateliers principle.

Answer:

[tex]m_{Fe}=23.0gFe[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]FeO+Mg\rightarrow Fe+MgO[/tex]

Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:

[tex]n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO[/tex]

Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:

[tex]n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO[/tex]

Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:

[tex]m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\ \\m_{Fe}=23.0gFe[/tex]

Regards.

Answer:

296.1g of Ag is the maximum amount of silver

Explanation:

A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:

Mass of solution:

Assuming a density of 1g/mL:

[tex]4.7L \frac{1000mL}{1L} \frac{1g}{mL} = 4700g[/tex]

If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:

4700L × (6.3g Ag / 100g) =

Answer:

2.3125g

Explanation:

Half-life referred to the time required for a quantity to reduce to half of its initial value, It used to calculate how unstable atoms undergo, or the period of time and atom can survive, radioactive decay.

Given:

t(1/2)= 9.25days

Initial mass of Thulium-167 = 48grams

We need to calculate the remaining amount after 37days.

Since we know that 1 half life = 9.25 days

Then 37 days means ( 37/9.25) half lives

37days means 4 half life

That means the 38grams of Thulium-167 will be halved by 4 times.

Then the ratio between the initial Amount and the amount remaining after 37 days can be calculated as. 0.5^(4)

= 37days × 0.5^(4)

= 2.3125g

the remaining amount of Thallium-167 after 37days is 2.3125g

Answer:

Manganese (II) ion, Mn²⁺

Explanation:

Hello,

In this case, given the overall reaction:

[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]

Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺

Best regards.

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,  

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,  

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.  

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,  

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,  

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,  

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,  

x as 0.115 and y as 0.165

Now the mass of AgCl will be,  

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,  

278.1 × 0.165 = 45.88 grams.  

Answer:

HPO₄⁻² predominates at pH 9.3

Explanation:

These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.

H₃PO₄ + H₂O   ⇄   H₂PO₄⁻   +   H₃O⁺   pKa 2.14

H₂PO₄⁻  +  H₂O   ⇄   HPO₄⁻²  +  H₃O⁺  pKa 6.86

HPO₄⁻²  +  H₂O   ⇄   PO₄⁻³   +   H₃O⁺  pKa 12.38

The H₂PO₄⁻  works as amphoterous, it can be a base and acid, according to these equilibriums.

H₂PO₄⁻ +  H₂O   ⇄   HPO₄⁻²  +  H₃O⁺

H₂PO₄⁻ +  H₂O   ⇄   H₃PO₄  +  OH⁻

pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:

HPO₄⁻ +  H₂O   ⇄   H₂PO₄⁻  +  OH⁻

HPO₄⁻²  +  H₂O   ⇄   PO₄⁻³   +   H₃O⁺

The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:

(6.86 + 12.38) /2 =  9.62

Above this pH, [PO₄⁻³] > [HPO₄⁻²].

In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]

Answer:

[tex]C_2=0.97M[/tex]

Explanation:

Hello,

In this case, for dilution process, we can notice that the initial moles remain the same once the dilution is completed, therefore, both concentration and volume change considering:

[tex]n_1=n_2\\\\V_1C_1=V_2C_2[/tex]

In such a way for the given final volume, the resulting concentration is noticed to be:

[tex]C_2=\frac{V_1C_1}{V_2} =\frac{10mL*2.5M}{25.8mL}\\ \\C_2=0.97M[/tex]

This is supported by the fact that the higher the volume the lower the concentration.

Best regards.

Answer:

BF3

Explanation:

Hybridization can be defined as the mixing of two or more atomic pure orbitals. ( s, p ,  and d) to produce two or more hybrid atomic orbitals that are similar and identical in shape and energy e.g sp,sp²,sp³ ,sp³d, sp³d². Usually , the central atom of a covalent molecules or ion undergoes hybridization.

in BF3; Boron is the central atom. Here, A 2s  electron is excited from the ground state of boron ( 1s²2s²2p¹) to one empty orbitals of 2p.

The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals tat are trigonally arranged in the plane in order to minimize repulsion . Each of the three hybrid orbitals overlaps with p-orbital of fluorine atom to form three bonds of equal strength and with bond angles of 120⁰.

                                  Energy                                                                  

B     ⇵  ║   ⇅  ║    ↑     ----------->    *B    ⇵ ║     ↑  ║     ↑  ║   ↑

       1s      2s         2p                              1s       2s           2p  

    Ground state                                        Excited State

The above shows an illustrative example of how electrons  move from the ground state to the excited state.

Question

Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is _____, and the boiling point of toluene is ____

Answer:

The boiling point of cyclohexanol is 81°C and the boiling point of toluene is 111°C.

Explanation:

The boiling point of cyclohexanol is 81°C and the boiling point of toluene is 111°C.

Therefore, the liquid collected first should be​​​​​ cyclohexane. .

Cyclohexane is collected first in the distillation to separate cyclohexane and toluene.

In the distillation process the mixture of cyclohexane and toluene is heated therefore vapourizing the subtance. Under the boiling reflux, the vapour phase become richer in the low boilng component ,cyclohexane. The cyclohexane vapours continue to condense first as compare to high boiling component ,toluene. Hence, cyclohexane is collected first.

Answer:

The correct answer is 62.5 %.

Explanation:

Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.  

Hence, the total pressure is 710+50 = 760 mmHg

According to Dalton's law of partial pressure,  

Partial pressure = mole fraction * total pressure

Mole fraction of water is,  

Partial pressure of water/Total pressure = 710/760 = 0.93

Similarly, the mole fraction of butyl butyrate is,  

Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07

Therefore, mole% of water is 0.93 * 100 = 93 %

For calculating mass%,  

Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)

The molecular mass of butyl-butyrate is 144 gram per mole

The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams

The mass percent of water will be,  

Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.  

Answer:

  C  The experiment shows that the red substance experienced a chemical change.

Explanation:

Apparently, adding heat caused the red substance to decompose into a gas and a metallic liquid. If it were simply a phase change, the original red substance could be expected to return when the temperature cooled. Because the substance apparently decomposed, it is clearly not an element. At no point in the experiment is there any evidence of a plasma being formed.

The observed decomposition is a chemical change.

Answer:

0.49 mol

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCI ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 12 g of Mg

The molar mass of Mg is 24.31 g/mol.

[tex]12g \times \frac{1mol}{24.31g} = 0.49mol[/tex]

Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.

Answer:

[tex]t=2.08s[/tex]

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

[tex]ln(\frac{[A]}{[A]_0} )=kt[/tex]

Thus, we compute the time as shown below:

[tex]t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s[/tex]

Best regards.

Answer:

Explanation:

0,44

Answer:

Tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide because they become soluble whereas copper (II) sulfide and bismuth (III) sulfide remain insoluble.

Explanation:

Sodium hydroxide is a  basic solution which is used as a precipitating agent for metallic ions in the laboratory.

When a solution containing a mixture of the sulfides of the Group II cations,  antimony (III), copper (II), tin  (IV), and bismuth (III), is made basic by the addition of a base such as sodium hydroxide or ammonium hydroxide, the  sulfide ion concentration will increase. The sulfides of antimony (III) and tin (IV) will then become  soluble because antimony (III) and tin (IV) form stable complexes with sulfide, which are soluble in  water, while the sulfides of copper (II) and bismuth (III) do not. The result is the dissolution of the antimony (III) sulfide  and tin (IV) sulfide, separating them from the copper (II) sulfide and the bismuth (III)  sulfide.

Sb₂S₃(s) + 3S²⁻(aq) ----> 2SbS₃³⁻(aq)

SnS₂(s) + S²⁻(aq) ----> SnS₃³⁻(aq)

Answer:

Gaseous

Explanation:

Gasses can move freely and do not form the shape of their containers

Liquids are more free than solids, but they conform to the shape of their container

Solids are not free

Answer: Therefore, the volume of 0.001 M KI is 61 ml.

Explanation:

The balanced chemical reaction is :

[tex]2KI+Cu(NO_3)_2\rightarrrow CuI_2+2KNO_3[/tex]

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]KI[/tex] solution = 0.100 M

[tex]V_1[/tex] = volume of  [tex]KI[/tex] solution = ?

[tex]M_2[/tex] = molarity of [tex]Cu(NO_3)_2[/tex] solution = 0.200 M

[tex]V_2[/tex] volume of [tex]Cu(NO_3)_2[/tex] solution = 15.25 ml

[tex]n_1[/tex] = valency of  [tex]KI[/tex] = 1

[tex]n_2[/tex] = valency of  [tex]Cu(NO_3)_2[/tex] = 2

Putting in the values we get:

[tex]1\times 0.100\times V_1=2\times 0.200\times 15.25[/tex]

[tex]V_1=61ml[/tex]

Therefore, the volume of 0.001 M KI is 61 ml.

The question is incomplete, the solute was not given.

Let the solute be K₂CrO₄ and the solvent be water

Complete Question should be like this:

The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.

Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.  

Pvap = ________atm

Answer:

Pvap (of water above the solution) = 0.0306 atm

Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

Explanation:

Given

volume of solution = 1 Litre = 1000 mL of the solution

density of the solution = 1.063 g/mL

concentration of the solution= 0.438M

temperature of the solution= 298 K

vapour pressure of pure water = 0.0313atm

Recall: density = mass/volume

∴mass of solution = volume x density

m = 1000 x 1.063 = 1063 g

To calculate the moles of K₂CrO₄ = volume x concentration

= 1 x 0.438 = 0.438 mol

Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g

Mass of water = mass  of solution - mass of K₂CrO₄

= 1063 - 85.055 = 977.945 g

moles of water = mass/molar mass

∴ moles of water = 977.945/18.02 = 54.27 mol

 Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution

moles of ions = 3 x moles of K₂CrO₄

= 3 x 0.438 = 1.314 mol

Vapor pressure of solution = mole fraction of water x vapor pressure of water  

= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm

Answer:

Mg, Sc, Ca

Explanation:

To figure out increasing atomic radii, we use Periodic Trends applied to the Elements of the Periodic Table to help us out. We know that the trend for atomic radii is increasing left and down. Since Ca is the furthest down and left of the 3, it has the largest atomic radius. Since Sc is next element to Ca, it would be the 2nd largest atomic radius of the 3. Since Mg is above Ca, it has the smallest atomic radius of the 3.

Answer:

Joseph Henry.

Explanation:

Hello,

In this case, we define the accuracy based on the degree of deviation a measure has regarding the the true or expected value. In this case, the lower the difference, the more accurate the procedure. In such a way, for the two students, we compute the difference between the expected value and the obtained value:

[tex]Mary=2.300g/mL-2.226g/mL=0.074g/mL[/tex][tex]Joseph=2.335g/mL-2.300mL=0.035g/mL[/tex]

In such a way, since the difference is lower in Joseph's measure, we can say he is the most accurate.

Best regards.