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3 years ago

Which reaction follows the laws of conservation of matter

1 answer:

5 0

It's the law! Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants.

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Find δs∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia:12n2(g) 32h2(g)→nh3(g)

Dima020 [189]

Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).

What is standard entropy?

The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."

Calculation:

Balancing the given reaction following-

1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)

ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]

Here S° = standard entropy of the system

Insert into the aforementioned equation all the typical entropy values found in the literature:

ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]

⇒ΔS° = - 99.4 J/K

Therefore, the standard entropy, ΔS° is -99.4 J/K.

Learn more about standard entropy here:

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5 0

1 year ago

Which elements have fewer than four dots in their electron dot diagrams

Alex787 [66]

Hydrogen, helium, lithium,

6 0

4 years ago

Read 2 more answers

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

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

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

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

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: 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 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago

For molecules with more than one ionizable groups, such as glycine, which of the following statements is true? For molecules wit

tester [92]

Answer: Option (a) is the correct answer.

Explanation:

The isoelectric point (pI) is actually the pH where a particular amino acid exists as a neutral molecule. This means that if pH equals pI then there will be no electric charge on the molecule.

Therefore, the statement at pH near the pI, nearly all the molecules carry no net charge, is true.

A buffer is composed of a solution of weak acid or base and salt of weak acid/base.So, when pH equals pI then there occurs no net charge. Hence, there will be no existence of conjugate acid-base pair.

Therefore, the statement when the pH is near the pI, the solution is near its maximum buffering ability, is false.

It is known that when pH is less than pI then there will be a net positive charge on the amino acid.

Hence, the statement at a pH near the pI, nearly all the molecules carry a net positive charge, is false.

When pH is greater than pI then amino acid will have a net negative charge. Hence, the statement at a pH near the pI, nearly all the molecules carry a net negative charge, is false.

Thus, we can conclude that out of the given options the statement at pH near the pI, nearly all the molecules carry no net charge, is true.

5 0

3 years ago

How many electrons can the 5th shell hold? 8 18 24 32 50

krok68 [10]

Each shell contains a fixed number of electrons. The general formula of the number of electrons can be hold in n-th shell is 2n². 5th shell can hold 2x5² = 50 number of electrons.

1st shell contains 2 X 1²= 2 number of electrons, 2nd shell contains 2 X 2²=8 number of electrons, 3rd shell contains 2 X 3²=18 number of electrons, 4th shell contains 2 X 4²=32 number of electrons, 5th shell contains 2 X 5²=50 number of electrons and so on.

5 0

4 years ago

Read 2 more answers