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erica [24]
3 years ago
11

13. Copper(II) chloride and silver acetate Acid/Base:

Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
8 0

Answer:

13. CuCl_2 \hspace{0.1cm}+ CH_3COOAg -> AgCl \hspace{0.1cm} + (CH_3COO)_2Cu

This is a double displacement reaction, not an acid-base reaction

14.HF\hspace{0.1cm}+Ba(OH)_2 ->\hspace{0.1cm}H_2O\hspace{0.1cm}+BaF_2

Ba(OH)_2 -> Base

HF -> Acid

15. HCl\hspace{0.1cm}+Mg(OH)_2->\hspace{0.1cm}H_2O\hspace{0.1cm}+MgCl_2

MgCl_2 -> Base

HCl -> Acid

16.CH_3COOH\hspace{0.1cm}+LiOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+CH_3COOLi

LiOH -> Base

CH_3COOH -> Acid

17.H_2SO_4\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaHSO_4

NaOH -> Base

H_2SO_4-> Acid

18. H_2SO_4\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaHSO_4

NaOH -> Base

H_2SO_4-> Acid

19.H_2SO_4\hspace{0.1cm}+NaHCO3->\hspace{0.1cm}H_2O\hspace{0.1cm}+CO_2+\hspace{0.1cm}Na_2SO_4

NaHCO_3-> Base

H_2SO_4-> Acid

20.H_3PO_4\hspace{0.1cm}+Zn(OH)_4->\hspace{0.1cm}H_2O\hspace{0.1cm}+Zn_3(PO_4)_4

Zn(OH)_4> Base

H_3PO_4-> Acid

21. HF\hspace{0.1cm}+Na_2SO_3->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaF\hspace{0.1cm}+\hspace{0.1cm}SO_2

NaSO_2> Base

HF-> Acid

22.HCl\hspace{0.1cm}+Na_2SO_3->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaCl\hspace{0.1cm}+\hspace{0.1cm}SO_2

NaSO_2> Base

HCl-> Acid

23.HCl\hspace{0.1cm}+Na_2S->\hspace{0.1cm}H_2S\hspace{0.1cm}+NaCl

Na_2S> Base

HCl-> Acid

24. NH_4Cl\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaCl+{0.1cm}NH_3

NaOH> Base

NH_4Cl-> Acid

Explanation:

All acid-base reaction have the same products, H_2O and a salt. Whit this in mind the acids would the compounds that produces the hydronium ion (H^+) and the bases would be the compounds that produces the hydroxide ion (OH^-)

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. Helium is stored at 293 K and 500 kPa in a 1-cm-thick, 2-m-inner-diamater spherical container made of fused silica. The area w
Ivahew [28]

Answer:  

(a) 45.17×10^-14 kg/s  

(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

Explanation:  

Helium gas at temperature T=293k  

Helium gas at pressure P= 500kPa  

The inner diameter of spherical tank is D_1 = 2m  

The inner radius of spherical tank is : r_1 = \frac{D_1}{2}  

= \frac{2}{2}  

=1m  

Thickness of the container r = 1cm =0.01m  

Outer radius of the spherical tank is ;  

t = r_2 - r_1  

-r_2 = -t - r-1  

multiplying through with (-) we have ;  

r_2 = t + r_1  

r_2 = 1 + 0.01m  

r_2 = 1.01m  

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as  

D_A_B =4.0 ×10^-14 \frac{m^2}{s}  

From table molar mass and gas constant, the molecular weight of helium is:

 

M = 4.003kg/kmol  

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.  

S_He = 0.00045 kmol/m³. bar  

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as  

C_B_I = S_H_e×P  

= 0.00045kmol/m³. bar × (5)  

C_B_I = 2.25×10^-3 kmol/m³  

From one dimensional mass transfer through spherical layers is expressed as:

N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}

substituting all the values in the above relation, we have;

M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}

M_di_f_f=11.42×10^-14kmol/s

(a) The mass flow rate is expressed as

M_di_f_f = MN_diff

M_di_f_f=4.003×11.42×10^-14

M_di_f_f=45.71×10^-14kg/s

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

N_di_ff =11.42×10^-14kmol/s

N_di_ff= 11.42×10^-14×7×24×3600 kmol/week

N_di_f_f=6.9×10^-8kmol/week

The volume of the spherical tank is V=\frac{4}{3} πr_1^3

V=\frac{4}{3}π×1^3

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week

N_di_f_f-final ≅ 0.86

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

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