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Afina-wow [57]
3 years ago
5

2C4H10 + 13 02 -- 8 CO2 + 10H20

Chemistry
2 answers:
Lorico [155]3 years ago
5 0

Answer:

We should use 0.55 moles C4H10 to produce 50 grams of H2O (option 1 is correct)

Explanation:

Step 1: data given

Mass of H2O = 50 grams

Molar mass H2O = 18.02 g/mol

Molar mass C4H10 = 58.12 g/mol

Step 2: The balanced equation

2C4H10 + 13 02 → 8 CO2 + 10H20

Step 3: Calculate moles H2O

Moles H2O = 50.0 grams / 18.02 g/mol

Moles H2O = 2.77 moles

Step 4: Calculate moles C4H10

For 2 moles C4H10 we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O

For 2.77 moles H2O we need 2.77/5 = 0.55 moles C4H10

We should use 0.55 moles C4H10 to produce 50 grams of H2O (option 1 is correct)

Anika [276]3 years ago
4 0

Answer:

Option 1. 0.55 mol of C₄H₁₀

Explanation:

The reaction is:

2C₄H₁₀ + 13O₂ → 8 CO₂ + 10H₂O

This is a combustion reaction where carbon dioxide and water are produced.

We convert mass of produced water to moles → 50 g / 18 g/mol = 2.78 moles of water.

The stoichometry states that:

10 moles of water are made by 2 moles of C₄H₁₀

Therefore 2.78 moles of water will be made by (2.78 . 2) / 10 = 0.55 mol of C₄H₁₀

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The acceleration is equal to force divided by mass which would be 100 / 50 which is 2 m/s2
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What is the number of protons in a neutral atom with 31 electrons and 38 neutrons?
PSYCHO15rus [73]

31

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A certain first-order reaction (a→products) has a rate constant of 9.60×10−3 s−1 at 45 ∘c. how many minutes does it take for the
Alexxandr [17]

The integrated rate law expression for a first order reaction is

ln\frac{[A_{0}]}{[A_{t}]}=kt

where

[A0]=100

[At]=6.25

[6.25% of 100 = 6.25]

k = 9.60X10⁻³s⁻¹

Putting values

ln\frac{100}{6.25}=9.6X10^{-3}t

taking log of 100/6.25

100/6.25 = 16

ln(16) = 2.7726

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7 0
3 years ago
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Calculate the osmotic pressure of a 5.5M glucose solution at 20 degrees Celsius
NeTakaya

The osmotic pressure is a colligative property which depends upon the number of molecules and not the type of molecules

The relation between osmotic pressure and concentration is

πV = nRT

where

π = Osmotic pressure [ unit atm] = ?

V = volume

n = moles

R = gas constant = 0.0821 L atm / mol K

T = temperature = 20°C = 20 + 273.15 K = 293.15 K

also

Molarity = moles / Volume

So

Molarity = n/V = 5.5 M

Putting values

π = MRT

π = 5.5 X 0.0821 X 293.15 = 132.37 atm

Osmotic pressure of given glucose solution will be 132.37 atm

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4 years ago
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed
KIM [24]

Answer:

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?

A) BrCH2CH2Br

B) CH3CH2CH2Br

C) CH3CHBr2

D) CH3CH2CH2CH3

E) BrCH2CH2CH2CH2Br

Explanation:

The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.

This is an example of free radical substitution.

The structure of ethane and its bromination is shown below:

Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).

Remaining all other products are possisble to form on free radical substitution of ethane.

8 0
3 years ago
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