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Maksim231197 [3]
3 years ago
11

Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many

grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O
Chemistry
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

Explanation:

Number of moles of propane:

=Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

= 44 * 6.8181

= 299.9964 grams

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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
A. Choose one of the four biogeochemical cycles that were discussed in this unit - carbon, nitrogen, water, or oxygen and descri
MAVERICK [17]

Water cycle simply refers to the continuous movement or flow of water from the atmosphere to the earth and back to the atmosphere

<h3>What is are the stages of water cycle?</h3>

These are the basic stages through which water passes. These stages are:

  • Evaporation
  • Condensation
  • Precipitation
  • Collection.

So therefore, water cycle simply refers to the continuous movement or flow of water from the atmosphere to the earth and back to the atmosphere

Learn more about water cycle:

brainly.com/question/25796102

#SPJ1

6 0
2 years ago
Calculate the concentration of acetate ion in a buffer solution made from 2.00 mL of 0.50 M acetic acid and 8.00 mL of 0.50 sodi
Lelu [443]

Answer:

1 M

Explanation:

Equation of reaction is;

CH3COOH + CH3COONa -------------------> 2CH3COO^- + NaH

1 moles of each of the reactants react to give 2 moles of the acetate ion.

From the question, we have that 2.00 mL that is (2÷1000)L of 0.50 M acetic acid reacted with 8.00 mL that is (8/1000)L of 0.50 sodium acetate.

Then from equation, n = CV -------------------------------------------(1).

Where n= number of moles, V= volume, C= concentration.

Number of moles,n of acetic acid = 0.50M× 2/1000L.

n(acetic acid)= 0.001 moles.

Number of moles,n of sodium acetate= 0.50M ×(8/1000)L.

n(sodium acetate)= 0.004 moles.

0.001 moles of acetic acid react with 0.004 moles of Sodium acetate

Therefore, acetic acid is the limiting reagent.

One mole of acetic acid produces 2 moles of acetate ion.

0.001 mole of acetic acid produces= 0.002 moles of acetate ion.

Using the equation (1) that is, n= CV.

0.002= C× 2/1000

C= 0.002/0.002

C= 1 M

8 0
3 years ago
Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9
EleoNora [17]

Answer : The current passing between the electrodes is, 1.056\times 10^{-2}A

Explanation :

First we have to calculate the charge of sodium ion.

q=ne

where,

q = charge of sodium ion

n = number of sodium ion = 2.68\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C

Now we have to calculate the charge of chlorine ion.

q'=ne

where,

q' = charge of chlorine ion

n = number of chlorine ion = 3.92\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C

Now we have to calculate the current passing between the electrodes.

I=\frac{q}{t}+\frac{q'}{t}

I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}

I=1.056\times 10^{-2}A

Thus, the current passing between the electrodes is, 1.056\times 10^{-2}A

4 0
3 years ago
I'm a really stuck in this. I kind of suck in chemistry, please help me. I would really appreciate it.
MArishka [77]
<span>Group 1 can be characterized as atoms that have 1 electron in their valence shell. This is valuable when dealing with these questions, because the loss or gain of valence electrons is what defines ionic relationships. When group 1 elements form ionic bonds with other atoms, they are extremely likely to lose their valence electron, since the nucleus has a weaker pull on it than, say, a chlorine atom has on its 7 valence electrons. The weaker pull between the nucleus and the valence electron of group 1 elements means that the radius is high, since the electron is more free to move with less pull on it. This also means that the first ionization energy is low, since it takes relatively little energy for that electron to be pulled away to another atom.</span>
3 0
3 years ago
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