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Vadim26 [7]
3 years ago
14

Non-electrolyte = A, 10.6 grams, dissolved in solvent B, 740 grams, the boiling point of the solution is higher than the boiling

point of the solvent
Pure dissolved in 0.284 ° C. Determine the Kb value of solvent B. Set the molecular weight of A = 106.
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

T_{solution}-T{solvent}=K_bm_solute

Considering that the difference in the temperature is 0.284°C, and the given molality by:

m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m

Now, solving for K_b, we get:

K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m

Best regards.

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(Please help will mark brainliest!)
arsen [322]

We know that Weighted atomic mass of Element is Calculated based upon their existence of isotopes and the Relative abundance of these isotopes.

Given that the Element X is Composed of :

Isotope                       Relative abundance                   Atomic Mass

⁵⁵X                                        70%                                               55

⁵⁶X                                        20%                                               56

⁵⁷X                                        10%                                                57

Weighted Atomic Mass of Element X :

= (0.70)(55) + (0.20)(56) + (0.10)(57)

= 38.50 + 11.2 + 5.70

= 55.4

So, the Weighted Atomic Mass of Element X is 55.4

6 0
3 years ago
A 6.40 g sample of a compound is burned to produce 8.37 g CO_2, 2.75 g H_2O, 1.06 g N_2, and 1.23 g SO_2. What is the empirical
LuckyWell [14K]

The empirical formula :

C₁₀H₁₆N₄SO₇

<h3>Further explanation</h3>

Given

6.4 g sample

Required

The empirical formula

Solution

mass C :

= 12/44 x 8.37 g

= 2.28

mass H :

= 2/18 x 2.75 g

= 0.305

mass N = 1.06

mass S :

= 32/64 x 1.23

= 0.615

mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g

Mol ratio :

= C : H : N : S : O

= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16

= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019

= 10 : 16 : 4 : 1 : 7

The empirical formula :

C₁₀H₁₆N₄SO₇

3 0
3 years ago
A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 351.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this
notka56 [123]
There is a specific formula to use for these type of problems. 

ln (P2/ P1)= Δvap/ R x (1/T1 - 1/T2)

R= 8.314
P1= 92.0 torr
T1= 23 C + 273= 296 K
P2= 351.0 torr
T2= 45.0 C + 273= 318 K

plug the values and solve for the unknown

ln( 351.0/ 92.0)= Δvap/ 8.314 x (1/296 - 1/318)

Δvap= 47630.6 joules

3 0
3 years ago
Read 2 more answers
If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
Fed [463]

Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

Mass of C₆H₁₄O₂ = 16.5 g

Moles of O₂ = 499 mol

Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

moles of C₆H₁₄O₂  = mass × molar mass

moles of C₆H₁₄O₂ =  16.5 g × 118 g/mol

moles of C₆H₁₄O₂ = 1947 mol

Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

5 0
4 years ago
Read 2 more answers
KFell Fe"(CN), + e + Nat → KNaFe'Fe(CN)6
Alinara [238K]

Answer:

Most common oxidation states: +2, +3

M.P. 1535º

B.P. 2750º

Density 7.87 g/cm3

Characteristics: Iron is a gray, moderately active metal.

Characteristic reactions of Fe²⁺ and Fe³⁺

The [Fe(H2O)6]3+ ion is colorless (or pale pink), but many solutions containing this ion are yellow or amber-colored because of hydrolysis. Iron in both oxidation states forms many complex ions.

Aqueous Ammonia

Aqueous ammonia reacts with Fe(II) ions to produce white gelatinous Fe(OH)2, which oxidizes to form red-brown Fe(OH)3:

Fe2+(aq)+2NH3(aq)+3H2O(l)↽−−⇀Fe(OH)2(s)+2NH+4(aq)(1)

Fe3appt.gif

Aqueous ammonia reacts with Fe(III) ions to produce red-brown Fe(OH)3:

Fe3+(aq)+3NH3(aq)+3H2O(l)↽−−⇀Fe(OH)3(s)+3NH+4(aq)(2)

Fe3bppt.gif

Both precipitates are insoluble in excess aqueous ammonia. Iron(II) hydroxide quickly oxidizes to Fe(OH)3 in the presence of air or other oxidizing agents.

Sodium Hydroxide

Sodium hydroxide also produces Fe(OH)2 and Fe(OH)3 from the corresponding oxidation states of iron in aqueous solution.

Fe2+(aq)+2OH−(aq)↽−−⇀Fe(OH)2(s)(3)

Fe4appt.gif

Fe3+(aq)+3OH−(aq)↽−−⇀Fe(OH)3(s)(4)

Fe4bppt.gif

Neither hydroxide precipitate dissolves in excess sodium hydroxide.

Potassium Ferrocyanide

Potassium ferrocyanide will react with Fe3+ solution to produce a dark blue precipitate called Prussian blue:

K+(aq)+Fe3+(aq)+[Fe(CN)6]4−(aq)↽−−⇀KFe[Fe(CN)6](s)(5)

Fe5a1ppt.gif

With Fe2+ solution, a white precipitate will be formed that will be converted to blue due to the oxidation by oxygen in air:

2Fe2+(aq)+[Fe(CN)6]4−(aq)↽−−⇀Fe2[Fe(CN)6](s)(6)

Fe5a2ppt.gif

Many metal ions form ferrocyanide precipitates, so potassium ferrocyanide is not a good reagent for separating metal ions. It is used more commonly as a confirmatory test.

Potassium Ferricyanide

Potassium ferricyanide will give a brown coloration but no precipitate with Fe3+. With Fe2+, a dark blue precipitate is formed. Although this precipitate is known as Turnbull's blue, it is identical with Prussian blue (from Equation 5).

K+(aq)+Fe+2(aq)+[Fe(CN)6]3−(aq)↽−−⇀KFe[Fe(CN)6](s)(7)

Fe5b.gif

Potassium Thiocyanate

KSCN will give a deep red coloration to solutions containing Fe3+:

Fe+3(aq)+NCS−(aq)↽−−⇀[FeNCS]+2(aq)(8)

Fe5cppt.gif

No Reaction

Cl−, SO2−4

7 0
3 years ago
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