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nalin [4]
3 years ago
9

A 425.00-gram sample of a compound decomposes into 196.01 grams of carbon, 41.14 grams of hydrogen, 130.56 grams of oxygen, and

57.29 grams of silicon. Experiments have shown the compound has a molecular weight of 208.329. What is the molecular formula?
A.
C8H5O2Si
B.
C2H5OSi
C.
C4H10O2Si
D.
C8H20O4Si
Chemistry
1 answer:
Alenkasestr [34]3 years ago
3 0

The molecular formula is D. C_8H_20O_4Si.

<em>Step 1</em>.Calculate the <em>empirical formula </em>

a) Calculate the moles of each element

Moles of C= 196.01 g C × (1 mol C/12.01 g C) = 16.325 mol C  

Moles of H = 41.14 g H × (1 mol H/1.008 g H) = 40.813 mol H

Moles of O = 130.56 g O × (1 mol O/16.00 g O) = 8.1650 mol O

Moles of Si = 57.29 g Si × (1 mol Si/28.085 g Si) = 2.0399 mol Si

b) Calculate the molar ratio of each element

Divide each number by the smallest number of moles and round off to an integer

C:H:O:Si = 8.0027:20.008:4.0027:1 ≈ 8:20:4:1

c) Write the empirical formula

EF = C_8H_20O_4Si

<em>Step </em>2. Calculate the <em>molecular formula</em>

EF Mass = 208.33 u

MF mass = 208.329  u

MF = (EF)_n

n = MF Mass/EF Mass = 208.329 u/208.33 u = 1.0000 ≈ 1

MF = C_8H_20O_4Si

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jeyben [28]
A density of a substance is constant. It is an extensive property, meaning it does not depend on the amount of substance because it is a ratio of mass to volume. No matter how much of each there is, they would always have a fixed ratio called density. For lead, the density is

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Density = 23.94 g/ 2.10 cm³
Density = 11.4 g/cm³
4 0
3 years ago
14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
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Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

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pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
3 years ago
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Answer : The volume of 6M NaOH stock solution is, 16.7 mL

Explanation :

To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of NaOH stock solution.

M_2\text{ and }V_2 are the molarity and volume of NaOH.

We are given:

M_1=6M\\V_1=?\\M_2=0.2M\\V_2=500mL

Putting values in above equation, we get:

6M\times V_1=0.2M\times 500mL\\\\V_1=16.7mL

Thus, the volume of 6M NaOH stock solution is, 16.7 mL

4 0
3 years ago
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Read 2 more answers
A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.05 nm. It then gives off a photon having a wave
vodka [1.7K]
Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m 

<span>Energy emitted
Ee = hc/ 1736^-9m </span>

Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>

<span>Converting J to eV (1.60^-19 J/eV)
 ∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>

<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>

<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>

<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>

n² = -13.60 / -0.85 = 16
n = 4
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3 years ago
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