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3 years ago

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The reaction for the burning of one type of match tip is as follows: P4S3(s) + 8 O2(g) P4O10(g) + 3 SO2(g) What volume of SO2 at

STP can be produced by burning 1.00 g P4S3 in excess O2?

1 answer:

miskamm [114]3 years ago

4 0

a] work out mass of 1 mole P4S3 -- call this Y grams

b] eqtn tells you 1 mole P4S3 gives 3 moles SO2 [= 3 x 22.4L at STP]

c] by simple proportion

1g P4S3 gives (3 x 22.4)/Y SO2 at STP answer

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What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO

docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

$\rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6$

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2$

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

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2 years ago

1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas

hram777 [196]

A 20 L sample of the gas contains 8.3 mol N₂.

According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant

<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁

<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁

___________

<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L

<em>n</em>₂ = ?; <em>V</em>₂ = 20 L

∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

4 0

3 years ago

Which of the following is not a correct chemical equation for a double displacement reaction? (D?)

emmasim [6.3K]

The correct answer would be the last option. A double displacement type of reaction involves the switching of places the cations and anions accordingly. The given reaction is erroneous since in the product side the anions and cations are being paired which would not make sense. The correct reaction should be

4NaBr + Co(SO3)2 yields <span>CoBr4 + 2Na2SO3</span>

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3 years ago

How many moles are in 20 grams of Ar?

pashok25 [27]

Answer:

moles Ar in 20g = 0.500 mole Ar

Explanation:

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7 0

2 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>

6 0

3 years ago