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4 years ago

Using the round 2 fasttrack bond table (page 2), how is digby's "yield" calculated for their 10.0s2020 bond?

1 answer:

Montano1993 [528]4 years ago

5 0

This is based on the round 2 FastTrack bond table, Digby’s “yield” for their 10.0S2020 bond is calculated in this format:

= ($100 - 93.41) / 100

= 10.0 / 93.41

= (2480 - 2020) / 2020

= (2480 - 2020) / 2480

= 0.1855 or 15.55% is the yield

NOTE: It isn’t computed, it is reported.

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Determine the empirical formula of a compound having the following percent composition by mass: K: 24.74%; Mn: 34.76%; O: 40.50%

Dovator [93]

Answer: The empirical formula of the compound is $KMnO_4$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Let the mass of the compound be 100 g

Given values:

% of K = 24.7%

% of Mn = 34.76%

% of O = 40.50%

Mass of K = 24.7 g

Mass of Mn = 34.76 g

Mass of O = 40.50 g

To calculate the empirical formula of a compound, few steps need to be followed:

Step 1: Calculating the number of moles of each element

We know:

Molar mass of K = 39.10 g/mol

Molar mass of Mn = 54.94 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol$

$\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol$

$\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol$

Step 2: Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles

$\text{Mole fraction of K}=\frac{0.632}{0.632}=1$

$\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1$

$\text{Mole fraction of O}=\frac{2.53}{0.632}=4$

Step 3: Writing the mole fraction as the subscripts of each of the element

The empirical formula of the compound becomes $K_1Mn_1O_4=KMnO_4$

Hence, the empirical formula of the compound is $KMnO_4$

4 0

3 years ago

Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three s

stiks02 [169]

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 = 0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x = 1.053 g / 0.0405 mol = 26 g/mol

7 0

4 years ago

What are some things that increase the carbon dioxide level in the atmosphere? Select all that apply

dimaraw [331]

Burning- just big heat, from bruning forest

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4 years ago

True or false? 11H+31H⟶42Be

hoa [83]

Answer:

✅

Explanation:

true

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2 years ago

Which statement best relates an emprical formula with a molecular formula

netineya [11]

The empirical formula of a substance is just the molecular formula simplified

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4 years ago