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2 years ago

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A generic gas, X , is placed in a sealed glass jar and decomposes to form gaseous Y and solid Z . 3 X ( g ) − ⇀ ↽ − Y ( g ) + Z

( s ) How are these equilibrium quantities affected by the initial amount of X ( g ) placed in the container? Assume constant temperature

1 answer:

Anarel [89]2 years ago

6 0

Answer:

3 X ( g ) ⇄ Y ( g ) + Z ( s )

According to le chatelier principle increases in reactant concentration will shift equilibrium towrads right direction.

$\frac{[Y]}{[X]^{3} }$ is constant with change in initial concentration of X.

Explanation:

3 X ( g ) ⇄ Y ( g ) + Z ( s )

Equilibrium constant for the reaction is as follows

$K_{c}$ = $\frac{[Y]}{[X]^{3} }$

At equilibrium there is no change in the ratio $\frac{[Y]}{[X]^{3} }$ of concentration of products to reactant if initial concentration change.

Therefore $\frac{[Y]}{[X]^{3} }$ is constant with change in initial concentration of X.

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A bird building a nest for its young is a reproductive strategy. Based on this example, what do you think the term “reproductive

Scilla [17]

A reproductive strategy is something done with the intent of preserving a family line so that life can continue.

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2 years ago

Read 2 more answers

Read the information on Hydrochloric Acid. List four ways you can avoid injury while using HCl and three steps you would take in

lisabon 2012 [21]

Answer:

Find the list and explanation below.

Explanation:

Hydrochloric acid shortened as HCl is a harmful chemical when not handled properly. Four ways to avoid injury with hydrochloric acid include;

1. Dilution: It is advised to dilute the concentrated hydrochloric acid with a reasonable amount of water so that its corrosive effect can be significantly reduced.

2. Wear a laboratory coat: When working with hydrochloric acid in the laboratory it is always advised to wear a laboratory suit so as to ensure a minimal amount of harm in case of an accident or spillage with the acid.

3. Wear a face shield: It is also important that a face shield is worn while working with the acid to prevent contact with the eyes.

4. Wear a protective rubber glove: The rubber gloves would ensure that there is minimal contact of the acid with the hands.

Three steps to be taken in case of an accidental exposure include;

1. Medical Attention: In serious cases, such as in a case of ingesting the acid, urgent medical help should be sought for immediately.

2. Water: If there is contact with the skin or eye, a lot of water should be used on the affected area, and then medical help obtained.

3. Inhalation of fresh air: In a case whereby the hydrochloric acid was inhaled for a long time, the affected person should try to get some fresh air and if other effects are noticed, urgent medical attention should be gotten.

3 0

2 years ago

What mass of NH3 can be made from 1221g of H2 and excess N2

expeople1 [14]

The mass of NH3 which can be made from 1221g of H2 and excess N2 is 6867g NH3

Explanation:

NH3 mass is 6867g
H2 mass is 1221g
N2 in excess

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2 years ago

Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00

AnnZ [28]

Answer:

Hello some parts of your question is missing

Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.

Answers:

a) 46.666.66 g/mol

b) 20930.23 g/mol

c)43333.33 g/mol

Explanation:

A)The equal number of molecules of each sample can be calculated using $Mn = \frac{n(M1 + M2 + M3)}{3n}$

because for the number of molecules to be equal : n1 = n2 = n3 = n

Mn = 46666.66 g/mol

B ) To calculate the equal masses of each sample

we apply this equation

$Mn = \frac{W1 + W2 +W3}{\frac{W1}{M1} +\frac{W2}{M2}+ \frac{W3}{M3} }$

ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION

7 0

2 years ago

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

<u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

<u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

4 0

2 years ago