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LuckyWell [14K]
3 years ago
15

SO2Cl2 decomposes in a first-order process with a half-life of 4.88 × 103 s. If the original concentration of SO2Cl2 is 0.011 M,

how many seconds will it take for the SO2Cl2 to reach 0.0032 M?
Chemistry
1 answer:
DiKsa [7]3 years ago
5 0
The amount of substance after n half-lives is calculated through the equation,
                                     A(n) = A(0) x (0.5)^t/h
where A(n) is the amount of n half-lives, A(0) is the original amount, t is the total time and h is the half-life. 
Substituting the known values,
 Basis is 1L,
                           0.0032 = (0.011) x (0.5)^(t/4.88x10^3s)
The value of t from the generated equation is equal to 8693.0 s or also equal to 8.693 x 10^3 s. 
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What’s a giant hunk of rock and metal
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5 0
3 years ago
An important industrial route to extremely pure acetic acid is the reaction of methanol with carbon monoxide:
Nastasia [14]

The reaction has had a heat that is enthalpy of -22 kJ/mol. The exothermic process has been signaled by the negative sign.

The amount of energy that the system absorbs or releases to create the products is described as the heat of reaction.

The source of the reaction's heat is

H is equal to 3(413 Kj/mol) + 358 Kj/mol + 467 Kj/mol + 1070 Kj/mol = 3134 Kj/mol.

H prod equals 3(413 kj/mol) plus 347 kj/mol plus 358 kj/mol plus 467 kj/mol plus 745 kj/mol, or 3156 kj/mol.

H=3134 kj/mol - 3156 kj/mol = -22 Kj/mol

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7 0
1 year ago
If you have 10 grams of Lithium Oxide what will the volume be? Show your work
gogolik [260]

Answer:

Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L

Explanation:

STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.

According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .

Molar mass of Lithium Oxide = 29.8 g/mol

Li_{2}O = 2(6.9) + 15.99 = 29.8

Mass of 1 mole Li_{2}O = 29.8 g

1 mole Li_{2}O occupies, Volume = 22.4 L

29.8 g Li_{2}O occupies, V = 22.4 L

1 g Li_{2}O occupies ,V = \frac{22.4}{29.8}

1 g Li_{2}O occupies ,V = 0.7516 L

10 g tex]Li_{2}O[/tex] occupies ,V = 0.7516 \times10 L

V = 7.52 L

So, volume occupied by Lithium Oxide At STP is 7.52 L

4 0
3 years ago
Five calcite, CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 12.2±0.1 g . What is the average mass and abs
Westkost [7]

Answer:

  • <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>

<em />

  • <em>The absolute uncertainty is: </em><u>0.008 g</u>

Explanation:

The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.

When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.

The relative uncertainty is the absolute uncertainty divided by the quantity:

  • Relative uncertainty = 0.1g / 12.2 g = 0.008

The average mass of calcium is calculated using proportions, along with the molar masses:

  • Molar mass of calcium: 40.078 g/ mol (from a periodic table)

  • Molar mass of calcite: 100.085 g/mol (given)

Proportion:

  • 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite

  • x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium

So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:

  • Average mass = total mass of five samples / number of samples

  • Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>

So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as  you have only used multiplication and division).

The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:

  • Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g

The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.

7 0
4 years ago
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