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UNO [17]
3 years ago
10

What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, C2H6? In this reaction,

ethane is burned in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O). ____C2H6(g)+____O2(g)→____CO2(g)+____H2O(g)
Chemistry
1 answer:
valina [46]3 years ago
5 0

<u>Answer:</u> The coefficients for balancing the given chemical equation are 2, 7, 4 and 6

<u>Explanation:</u>

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another form. This also means that total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

The given balanced chemical equation follows:

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

<u>On reactant side:</u>

Number of carbon atoms = 4

Number of hydrogen atoms = 12

Number of oxygen atoms = 14

<u>On product side:</u>

Number of carbon atoms = 4

Number of hydrogen atoms = 12

Number of oxygen atoms = 14

Hence, the coefficients for balancing the given chemical equation are 2, 7, 4 and 6

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Brrunno [24]
A. immigration, how is this chemistry?
8 0
3 years ago
A reaction is occurring in a test tube. How is heat transmitted to the surroundings?
Cerrena [4.2K]

It should be noted that when a  reaction is occurring in a test tube, heat transmitted to the surroundings when Molecules collide with the glass, and the glass molecules then transmit that energy to the outside.

<h3>What is heat?</h3>

Heat can be regarded as a form of energy which is energy that is been transferred as a result of difference in temperature.

In the case above, Molecules collide with the glass, and the glass molecules then transmit that energy to the outside which is an exothermic reaction.

Therefore, option B is correct.

Learn more about heat at:

brainly.com/question/12072129

8 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

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

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52
Andru [333]

Answer:

101,37°C

Explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m <em>(1)</em>

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>

<em />

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

<em>X = 101,37°C</em>

<em></em>

I hope it helps!

7 0
3 years ago
How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

<u>Step 2: Identify Conversions</u>

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.015733 g Au ≈ 0.0157 g Au

3 0
3 years ago
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