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DiKsa [7]
3 years ago
6

The balanced equation for a hypothetical reaction is a + 5b + 6c → 3d + 3e. what is the rate law for this reaction? rate = k [a]

[b]5[c]6 rate = k [a][b][c] rate = k [d]3[e]3
Chemistry
1 answer:
nekit [7.7K]3 years ago
7 0

<u>Answer: </u>The correct rate of the reaction is Rate=k[a][b]^5[c]^6

<u>Explanation:</u>

Rate law of the reaction is the expression which expresses the rate of the reaction in the terms of the molar concentrations of the reactants with each term raised to the power of their respective stoichiometric coefficients in a balanced chemical equation.

For the given reaction:

a+5b+6c\rightarrow 3d+3e

The expression for the rate law will be: Rate=k[a][b]^5[c]^6

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The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mariulka [41]

Answer:

3.09kg

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 from the balanced equation = 2 x 114 = 228g

Converting 228g of C8H18 to kg, we obtained:

228/1000 = 0.228kg

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Converting 704g of CO2 to kg, we obtained:

704/1000 = 0.704kg

From the equation,

0.228kg of C8H18 produced 0.704kg of CO2.

Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2

6 0
3 years ago
A piece of iron is heated with a torch to a temperature of 603 k.
Mazyrski [523]
This question asks to compare the energy emitted by a piece of iron at T = 603K with the energy emitted by the same piece at T = 298K.

Then you need to use the Stefan–Boltzmann Law

That law states that energy emitted (E) is proportional to fourth power of the  to the absolute temperature (T), this is E α T^4 (the sign α is used to express proportionallity.

Then E (603) / E (298) = [603K / 298K]^4 = 16,8

Which meand that the Energy emitted at 603 K is 16,8 times the energy emitted at 298K.


5 0
3 years ago
Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4
Katena32 [7]

<u>Answer:</u> The mass of rust that can be removed is 1.597 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = 6.00\times 10^2mL = 600 mL = 0.600 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol

For the given chemical reaction:

Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = \frac{1}{6}\times 0.06=0.01mol of ferric oxide (rust)

To calculate the mass of rust for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g

Hence, the mass of rust that can be removed is 1.597 grams

5 0
3 years ago
If a normal force of 500 N is acting on two surfaces in contact with a coefficient of friction equal to 0.50, what would be the
Makovka662 [10]

Answer:

The answer to this is

The maximum force of static friction is 250 N

Explanation:

To solve this, we least out the given variables thus

normal force, N = 500 N

coefficient of friction, μ  = 0.50

the friction formula is F = μ × N

Therefore the maximum force of static friction will be

0.50 × 500 N = 250 N

Force due to friction is the resisting force to the relative motion of two contacting surfaces, tending to prevent them from sliding.

The force that describes the resistance to the relative motion of two contacting solid surfaces is dry friction force

7 0
3 years ago
What volume of water has the same mass as 150 cm3 of gold?
MA_775_DIABLO [31]

Density of gold is 19.32 \frac{g}{cm^{3}}

Given volume of gold = 150cm^{3}

Calculating the mass of gold from volume and density:

Density =\frac{Mass}{Volume}

Mass = Density * Volume

= \frac{19.32g}{cm^{3} }  *150cm^{3}  = 2898 g

Density of water = 1\frac{g}{cm^{3} }

Calculating the volume of 2898 g water:

2898 g *\frac{1cm^{3} }{1 g}   = 2898 cm^{3}

Therefore, 2898cm^{3} water has the same mass as gold of volume 150 cm^{3}

3 0
3 years ago
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