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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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Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

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7 0
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Which of the following molecules experience dipole-dipole forces as its strongest IMF? A) H2 B) SO2 C) NH3 D) CF4 E) BCl3
Xelga [282]

Answer: NH_3

Explanation:

a) H_2: This  is a non polar covalent compound which are held by weak vanderwaal forces of attraction.

b) SO_2: This  is a covalent compound which is polar due to the presence of lone pair of electrons and are held by dipole-dipole forces of attraction.

c) NH_3: These are joined by a special type of dipole dipole attraction called as hydrogen bond. It forms between electronegative nitrogen atom and hydrogen atom and is the strongest interaction.

d) CF_4: This  is a covalent compound and is non polar which are held by weak vanderwaal forces of attraction.

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5 0
3 years ago
he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp
wolverine [178]

1) Drift velocity: 3.32\cdot 10^{-4}m/s

2) 5.6\cdot 10^{13} electrons per person

Explanation:

1)

For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:

v_d=\frac{I}{neA}

where

I is the current

n is the concentration of free electrons

e=1.6\cdot 10^{-19}C is the electron charge

A is the cross-sectional area of the wire

The cross-sectional area can be written as

A=\pi r^2

where r is the radius of the wire. So the equation becomes

v_d=\frac{I}{ne\pi r^2}

In this problem, we have:

I = 8.0 A is the current

8.5\cdot 10^{28} m^{-3} is the concentration of free electrons

d = 1.5 mm is the diameter, so the radius is

r = 1.5/2 = 0.75 mm = 0.75\cdot 10^{-3}m

Therefore, the drift velocity is:

v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s

2)

The total length of the cord in this problem is

L = 3.00 m

While the cross-sectional area is

A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2

Therefore, the volume of the cord is

V=AL (1)

The number of electrons per unit volume is n, so the total number of electrons in this cord is

N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}

In total, the Earth population consists of 8 billion people, which is

N'=8\cdot 10^9

Therefore, the number of electrons that each person would get is:

N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}

7 0
3 years ago
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