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3 years ago

Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?

Chemistry

1 answer:

charle [14.2K]3 years ago

8 0

Answer:

The answer is " $41.23 \ L\ N_2$ "

Explanation:

$2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049 \frac{\ g NO_3^{-}}{mol})} \times \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\$

$=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23$

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Do anyone know how to do question B

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a) IUPAC Names:

1) (trans)-but-2-ene

2) (cis)-but-2-ene

3) but-1-ene

b) Balance Equation:

C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄

As H₃PO₄ is catalyst and remains unchanged so we can also write as,

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c) Rule:

When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.

d) C is not Geometrical Isomer:

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2 years ago

What is true for two pieces of iron at the same temperature?

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3 years ago

Read 2 more answers

Balance the following: C₂ + B → B₂C₂ Ag₄ + SO₂ → AgO + S₂

jarptica [38.1K]

Answer:

The balanced reactions are:

C₂ + 2B → B₂C₂

Ag₄ + 2SO₂ → 4AgO + S₂

Explanation:

The Law of Conservation of Matter postulates that "the mass is not created or destroyed, only transformed." This means that the reagents interact with each other and form new products with physical and chemical properties different from those of the reagents because the atoms of the substances are ordered differently. But the amount of matter or mass before and after a transformation (chemical reaction) is always the same, that is, the quantities of the masses involved in a given reaction must be constant at all times, not changing in their proportions when the reaction ends. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

In other words, the law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts .

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

You have:

C₂ + B → B₂C₂

Left side: 2 C and 1 B.

Right side: 2 C and 2 B.

Carbon C is balanced since you have the same amount of the element on each side of the reaction, but boron B is not balanced. So, balancing the reaction:

C₂ + 2B → B₂C₂

Left side: 2 C and 2 B.

Right side: 2 C and 2 B.

Now you have the same amount of each element on each side of the reaction, so the reaction is balanced.

Now you have:

Ag₄ + SO₂ → AgO + S₂

Left side: 4 Ag, 1 S and 2 O.

Right side: 1 Ag, 2 S and 1 O.

Balancing the Ag:

Ag₄ + SO₂ → 4AgO + S₂

Left side: 4 Ag, 1 S and 2 O.

Right side: 4 Ag, 2 S and 4 O.

Balancing the O:

Ag₄ + 2SO₂ → 4AgO + S₂

Left side: 4 Ag, 2 S and 4 O.

Right side: 4 Ag, 2 S and 4 O.

You have the same amount of each element on each side of the reaction, so the reaction is balanced.

3 0

3 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

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 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$