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3 years ago

Consider the chemical equation.

Chemistry

2 answers:

frez [133]3 years ago

8 0

NBr₃ is the answer young man.

likoan [24]3 years ago

5 0

NBr₃ : NaOH = 2 : 3
40 : 48 = 2 : 2.4
40 : 48 = 2.5 : 3

NBr₃ is the excess reactant

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How many moles of CH4 are in 148.6 g? Round your answer to two decimal places.

Dafna1 [17]

Answer:

9.29 mol

Explanation:

Given data:

Number of moles = ?

Mass = 148.6 g

Solution:

Number of moles = mass/ molar mass

Molar mass of CH₄ = 16 g/mol

Now we will put the values in formula.

Number of moles = 148.6 g/ 16 g/mol

Number of moles = 9.29 mol

Thus 148.6 g have 9.29 moles.

6 0

3 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

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 $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

Make a prediction about the relationship between electrons and molecular shapes

Oksana_A [137]

Answer:

★ Molecular geometry is described by VSEPR theory, which basically states that electron pairs around a central atom will repel each other, and get as far apart as possible, in three dimensions.

Explanation:

Hope you have a great day :)

6 0

3 years ago

There are four substances, A, B C and D. A and C are pure substances. They cannot be broken down into anything simpler by chemic

yarga [219]

The answer is C Becuz D is a compound and that is the only answer that says it

6 0

3 years ago