1 mole of alkene CxH2x was fully burnt in oxygen. The products were analysed. 264g of Co2 and 108g of of H20 were produced. Use the information to balance the equation and work out the identity of CxH2x.
CxH2x+ O2--> CO2+H2O
PLEASE CAN SOMEONE EXPLAIN THIS TO ME!!!!!!!!!!!!

Answers

Answer 1

Answer:

C6H12

Explanation:

Step 1: Find the molar mass of carbon dioxide and water

MH2O = 2(1.008) + 16.00 = 1.802x10^1 g/mol

MCO2 = 12.01 + 2(16.00) = 4.401x10^1 g/mol

Step 2: Calculate the moles of the products

nH2O = 108g / 1.802x10^1 g/mol = 5.99 or about 6

nCO2 = 264g/ 4.401x10^1 g/mol = 5.99 or about 6

Step 3: Enter moles of carbon dioxide and water into the balanced equation

CxH2x + O2 = 6CO2 + 6H2O

Step 4: Balance

We see how there is six carbon dioxide on the right side which means there are six carbons in the equation.

This means x is equal to 6 in our equation.

If you plug the information into the equation you get:

C6H12 + O2 = 6 CO2 + 6 H2O

Now all that's left is to balance the oxygens

We see how there is 18 oxygens on the right side of the equation which means there must be 18 on the left side.

Because we have oxygen gas we divide 18 by 2 which means there are 9 O2's on the left side

Therefore, the balanced equation is C6H12 + 9O2 = 6 CO2 + 6 H2O


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Answer:

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Question 22 of 30
What is the balanced form of the chemical equation shown below?
C6H1206(s) + O2(g) → H2O(l) + CO2(9)
A. CH 0(s) + O2(g) → H2O(1) + CO2(g)
B. CoH1206(s) + O2(g) → 12H2O(l) + 6CO2(g)
C. C6H1206(s) + O2(g) → H20(1) + CO2(g)
D. CoH 206(s) + 602(g) → 6H2O(l) + 6CO2(g)
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Answers

Answer:

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Explanation:

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3-bakelite
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Answers

Answer:

b. stainless steel

Explanation:

b) stainless steel is not good condoctor of heat

True or false, The universal gas constant, R, is . 0821 when we use mm Hg as our unit for pressure

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Answer:

This is true.

Explanation:

R should be 0.0821

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Answers

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Answer:

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Answer:

Explanation:

Element -

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Answers

Answer:

The first step in the Scientific Method is to make objective observations. These observations are based on specific events that have already happened and can be verified by others as true or false. Step 2. Form a hypothesis.

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Answer:

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Answer:

B

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A 45g Aluminum spoon (specific heat 0.88J/g degree Celcius) at 24 degrees Celcius placed in 180ml(180g) of coffee at 85 degrees Celcius and the temperature of the two become equal. What is the final temperature?

Answers

Answer:

82 °C

Explanation:

Let the specific heat capacity of the coffee be that of water which is 4.2 J/g °C.

Now, at the final temperature, heat gained by Aluminum spoon ,Q equals heat lost by coffee, Q'.

Q = -Q'

Q = m₁c₁(T₂ - T₁) where m₁ = mass of aluminum spoon = 45 g, c₁ = specific heat of aluminum = 0.88 J/g °C, T₁ = initial temperature of aluminum spoon = 24 °C and T₂ = final temperature of aluminum spoon.

Q' = m₂c₂(T₂ - T₃) where m₂ = mass of coffee = 180 g, c₂ = specific heat of coffee = 4.2 J/g °C, T₃ = initial temperature of coffee = 85 °C and T₂ = final temperature of coffee.

So, Q = -Q'

m₁c₁(T₂ - T₁) = -m₂c₂(T₂ - T₃)

Making T₂ subject of the formula, we have

m₁c₁T₂ - m₁c₁T₁ = -m₂c₂T₂ + m₂c₂T₃

m₁c₁T₂ + m₂c₂T₂ =  m₂c₂T₃ + m₁c₁T₁

(m₁c₁ + m₂c₂)T₂ =  m₂c₂T₃ + m₁c₁T₁

T₂ =  (m₂c₂T₃ + m₁c₁T₁)/(m₁c₁ + m₂c₂)

substituting the values of the variables into the equation, we have

T₂ =  (180 g × 4.2 J/g °C × 85 °C + 45 g × 0.88 J/g °C × 24 °C )/(45 g × 0.88 J/g °C  + 180 g × 4.2 J/g °C)

T₂ =  (64260 J + 950.4 J)/(39.6 J/°C  + 756 J/°C)

T₂ =  65210.4 J/795.6 J/°C

T₂ =  81.96 °C

T₂ ≅  82 °C

Name the property of salt investigated in the above experiment

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which experiment please involve a clip

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