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How many moles of LiOH are needed to react completely with 25.5 g of CO2

LekaFEV [45]

Answer:

3.18 mol

Explanation:

$2LiOH+CO_{2}-> Li_{2}CO_{3} +H_{2}O$

n(CO2) = mass/ Mr.

= 25.5 / 16

= 1.59 mol

As per the equation above,

n(LiOH) : n(CO2)

2 : 1

∴ 3.18 : 1.59

3 0

2 years ago

What is the name of the pores, or holes, on the outer layer of the sponge that take water in?

ivanzaharov [21]

The tiny holes in a sponges outer layer that take in water are called ostia, I believe. Hope it helps

7 0

3 years ago

Which statement describes the differences between chemical reactions and nuclear decay rates? Chemical reaction rates vary, but

DanielleElmas [232]

Answer:

Answer the last one Nuclear decay rates vary, but chemical reaction rates are constant

Explanation:

Correct me if im wrong

6 0

2 years ago

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What is the percent of Cr in Cr2O3

IgorLugansk [536]

The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
= $\frac{103.98}{151.98}$ × 100

= 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.

5 0

2 years ago

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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