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Vilka [71]
2 years ago
10

Given the balanced chemical equation, SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = −184 kJ Determine the mass (in grams) of H

F(g) must react when 345 kJ of energy is released.
Chemistry
1 answer:
Ostrovityanka [42]2 years ago
3 0

Answer:

mass HF = 150.05 g

Explanation:

  • SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

⇒ Q = (ΔH°rxn * mHF) / (mol HF * MwHF )

∴ MwHF = 20.0063 g/mol

∴ mol HF = 4 mol

∴ ΔH°rxn = - 184 KJ

∴ Q = 345 KJ

mass HF ( mHF ):

⇒ mHF = ( Q * mol HF * MwHF ) / ΔH°rxn

⇒ mHF = ( 345 KJ * 4mol HF * 20.0063 g/mol ) / 184 KJ

⇒ mHF = 150.05 g HF

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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate? C
Lubov Fominskaja [6]

Answer:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

Explanation:

For these reactions, the products are:

1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl

2. Li₂S(aq) + Pb(NO₃)₂(aq) →  PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

AgCl is an insoluble white compound.

As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

6 0
2 years ago
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

8 0
2 years ago
Choose chlorous acid, hclo2, from the list above and press the equilibrate button. when the 0.10 m solution of hclo2 is allowed
Arturiano [62]
Chlorous Acid Ionizes as,

                                        HClO₂   ⇆   H⁺  +   ClO₂⁻

                                        Ka  =  [H⁺] [ClO₂⁻] / [HClO₂]

 Ka of Chlorous Acid  =  1.1 × 10⁻²

The concentration of H⁺ ions at equilibrium are calculated as,

Initial                               0.1 M     ⇆    0          0
At Equilibrium                 0.1-X      ⇆    X         X
So,
                                       Ka  =  [X] [X] / [0.1-X]
Putting value of Ka,
                                1.1 × 10⁻²  =  X² / 0.1-X
Solving for X,
(www.cymath.com)
                                 X  =  0.028 M

Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.

Percentage Ionization is calculated as,

                                 =  [H⁺] / [HA] × 100

                                 =  (0.028 / 0.1) × 100

                                 =  0.28 × 100

                                 =  28 %     (Percentage Ionization)
6 0
2 years ago
A state property X has a value 89.6 units. It undergoes the following changes, first increase by
yanalaym [24]

Answer:

B) -4.1 units

Explanation:

According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:

- first increase by 3.6 units

- then increase by another 18.7 units

- then decrease by 12.2 units

- and finally attains a value of 85.5 units

This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5

To get x, we say;

89.6 + 3.6 = 93.2

93.2 + 18.7 = 111.9

111.9 - 12.2 = 99.7

99.7 - 85.5 = 14.2units.

The changes that occured is represented as follows:

= (3.6 + 18.7) - (12.2 + 14.2)

= 22.3 - 26.4

= -4.1 units

7 0
2 years ago
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