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3 years ago

If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant

Chemistry

1 answer:

Vanyuwa [196]3 years ago

6 0

<h3>Answer:</h3>

Excess Reagent = NBr₃

<h3>Solution:</h3>

The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,

2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO

Calculating the Limiting Reagent,

According to Balance equation,

2 moles NBr₃ reacts with = 3 moles of NaOH

So,

40 moles of NBr₃ will react with = X moles of NaOH

Solving for X,

X = (40 mol × 3 mol) ÷ 2 mol

X = 60 mol of NaOH

It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.

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Answer:

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Explanation:

8 0

3 years ago

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.2

alexandr402 [8]

Answer:

1) the wavelength of the next line in the series is 397.2 nm

2) The ionization energy is 3.3996 eV

Explanation:

Step 1: Data given

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.29 nm

Step 2: Calculate n₂

The visible spectrum of light from hydrogen displays four wavelengths, 410 nm, 434 nm, 486 nm, and 656 nm,

1/λ = Rh *(1/n₁² - 1/n₂² )

⇒with λ = the wavelength

⇒with Rh = Rydberg constant for hydrogen, 1.09677583 * 10^7 m

⇒ with n₁ = the principal quantum number of an energy level

⇒with n₂ = the principal quantum number of an energy level for the atomic electron transition

λ * Rh = n₁²* (n₁+1)² / (2n₁² + 1)

656.46 nm * 109677 cm = n₁²* (n₁+1)² / (2n₁² + 1)

7.20 = n₁²* (n₁+1)² / (2n₁² + 1)

n1 = 2

All those are in the visible spectrum and are called Balmer series, or Balmer lines.

n1 (the principal quantum number of an energy level) for Balmer series is: n1 = 2

Step 3: calculate he wavelength of the next line in the series?

1/λ = Rh *(1/n₁² - 1/n₂² )

⇒with n₂ = the principal quantum number of an energy level for the atomic electron transition = 7

1/λ = 109677.6 / cm * (1/2² - 1/7²)

1/λ = 109677.6 / cm * (1/4 - 1/51.84)

λ = 397.2 nm

the wavelength of the next line in the series is 397.2 nm

Step 4: What is the ionization energy of the atom when it is in the lower state of the transitions?

The energy required to ionize the atom is:

n₂ → ∞

V∞ = 1/λ = 109677.6 / cm * (1/4 - 0)

V∞ = 109677.6 * 1 eV/ 8065.5 cm-1

V∞ = 27419.25 * 1 eV / 8065.5 cm-1

V∞ = 3.3996 eV

The ionization energy is 3.3996 eV

8 0

4 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

<u>Answer:</u> The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

<u>Explanation:</u>

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

<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 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

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

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

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 = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

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

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

Which of the following happens when a reaction reaches dynamic equilibrium in a closed system?

laiz [17]

Answer: B

Explanation:

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus the concentrations of the reactants and products must be constant.

8 0

2 years ago

Susan needs to simplify the expression shown below what should be her first step

Digiron [165]

Susan should follow PEMDAS,

Parentheses
Exponents
Multiplication
Division
Addition
Subtraction,

So, the first step should be, to solve the equation in the parentheses.

I hope this helps!

5 0

4 years ago