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2 years ago

How many moles of KCl will be produced from 15.0 g of KClO3?

Chemistry

2 answers:

slega [8]2 years ago

7 0

Answer:

0.12 mol KCl

Explanation:

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

15 g x mol

x g KCl = 15 g KClO3 x[ (1 mol KClO3)/ (122.5 g KClO3) ] x [(2 mol KCl)/ (2 mol KClO3)]

x g KCl = 0.12 mol KCl

MAXImum [283]2 years ago

7 0

Answer:

0.1224 moles

Explanation:

this is a decompostion reaction

Equation of reaction

2KClO₃ → 2KCl + 3O₂

molar mass of KClO₃ =

molar mass of K = 39.09g/mol

Molar mass of Cl = 35.5g/mol

Molar mass of O = 15.99g/mol

Molar mass of KClO3 = [39 + 35.5 + (3*15.99)]

Molar mass of KClO3 = 122.47g/mol

From the equation of reaction,

2 moles of KClO₃ produces 2 moles of KClO₃

But 1 mole = molar mass

1 mole of KClO₃= 122.47g

2 moles of KClO₃ = 2 * 122.47 = 244.94g

If 244.94g of KClO₃ produces 2 moles of KCl

15 g of KClO₃ will produce x mole of KCl

x = (15 * 2) / 244.94

x = 0.1224moles

15g of KClO₃ will produce 0.1224moles of KCl

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How many grams of CO2 will be produced from 12.0 g of CH4 and 133 g of O2?

jek_recluse [69]

Answer:

145 grams

Explanation:

conservation of mass

12.0+133=145

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2 years ago

Calcium hydroxide, Ca(OH)2, is used as a calcium nutritional supplement in some foods and beverages, such as orange juice. What

RSB [31]

Answer:

$pH=11.08$

Explanation:

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In this case, since calcium hydroxide a strong base, its dissociation will completely result in both calcium and hydroxyl ions:

$Ca(OH)_2\rightarrow Ca^{2+}+OH^-$

Thus, the concentration of hydroxyl ions equals that of the calcium hydroxide, with which we could compute the pOH as shown below:

$pOH=-log([OH^-]}=-log(0.0012)\\\\pOH=2.92$

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Hence, the pH finally results:

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2 years ago

The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =

Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) (1)

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] (2)

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) = 9.744g of monosodium succinate.

Mass of disodium succinate:

0.0304moles * (162g / 1mol) = 4.925g of disodium succinate.

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2 years ago

Boron - 10 emits alpha particles and cesium - 137 emits beta particles. Write balanced nuclear

anzhelika [568]

Answer:

B10 5N +5P= Li6 3N +3P

Cs 137 82N+55P = I 133 80N + 53P

Explanation:

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2 years ago

Read 2 more answers

A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3

Sati [7]

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C

(58 g . 0.100) / 0.580 = 10 g of H

48 g of C / 12 g/mol = 4 mol

10 g of H / 1g/mol = 10 moles

7 0

2 years ago