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zepelin [54]
2 years ago
11

Why does Australian get more uv light than othet parts of the world?​

Chemistry
1 answer:
STALIN [3.7K]2 years ago
7 0
I definitely agree whithe the first person that the earths orbit brings that part closer to the sun
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Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric aci
Pavlova-9 [17]

Given:

K = 0.71 = Kp

The reaction of sulphur with oxygen is

                            S(s)   + O2(g)  ---> SO2(g)

initial Pressure                   6.90         0

Change                                -x            +x

Equilibrium                     6.90-x          x

Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)

4.899 - 0.71x  = x

4.899 = 1.71x

x = 2.86 atm = pressure of SO2 formed

temperature = 950 C = 950 + 273.15 K = 1223.15 K

Volume = 50 L

Let us calculate moles of SO2 formed using ideal gas equation as

PV = nRT

R = gas constant = 0.0821 L atm / mol K

putting other values

n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles

Moles of Sulphur required = 1.42 moles

Mass of sulphur required or consumed = moles X atomic mass of sulphur

mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg  of sulphur



 


6 0
3 years ago
67.4 L of sulfur dioxide (SO2) gas at 0.0C and at 1 atm of pressure is equivalent aren’t to how many moles?
gizmo_the_mogwai [7]

Solution here,

Volume(V)=67.4 L

Pressure(P)=1 atm

Temperature(T)=(0+273)K=273K

Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1

No. of moles(n)=?

Now,

PV=nRT

or, 1×67.4=n×0.0821×273

or, 67.4=22.4n

or, n=67.4/22.4

or, n=3

therefore, required no. of mole is 3.

6 0
3 years ago
A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f
Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0
2 years ago
List the ions in solution in the electrolysis of conc.sodium chloride<br>​
Licemer1 [7]
Chloride ions Cl –(aq) (from the dissolved sodium chloride) are discharged at the positive electrode as chlorine gas, Cl 2(g) sodium ions Na +(aq) (from the dissolved sodium chloride) and hydroxide ions OH –(aq) (from the water) stay behind - they form sodium hydroxide solution, NaOH(aq)
5 0
2 years ago
What group of substances does this substance belong to?
HACTEHA [7]
In comparison see it is very easy in goolge
4 0
3 years ago
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