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3 years ago

Why does his teacher ask him to balance the equation by including the correct coefficients?

Chemistry

1 answer:

goldenfox [79]3 years ago

8 0

Because correct coefficients are necessary to successfully balance equation.

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The process of fermentation involves chemical reactions that convert solid glucose (C6H12O6) into aqueous ethanol and carbon dio

e-lub [12.9K]

Answer:

C6H12O6 —> 2C2H5OH + 2CO2

Explanation:

The equation for the reaction is given below:

C6H12O6 —> C2H5OH + CO2

We can balance the equation above as follow:

There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:

C6H12O6 —> 2C2H5OH + CO2

There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C6H12O6 —> 2C2H5OH + 2CO2

Now the equation is balanced.

4 0

3 years ago

60.0 g of NH Cl are dissolved in 100 grams of water at 60.0° C. This solution is

Sidana [21]

Answer:

B) is the correct answer

3 0

3 years ago

When sugar dissolves in water are covalent bonds broken?

Amanda [17]

Gle's cache of http://www.middleschoolchemistry.com/lessonplans/chapter5/lesson4<span>. It is a snapshot of the page as it appeared on 21 Oct 2017 07:24:57 GMT.</span>

7 0

3 years ago

Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio

umka21 [38]

Answer : The value of $K_p$ at this temperature is 66.7

Explanation : Given,

Pressure of $PCl_3$ at equilibrium = 0.348 atm

Pressure of $Cl_2$ at equilibrium = 0.441 atm

Pressure of $PCl_5$ at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}$

Now put all the values in this expression, we get :

$K_p=\frac{(10.24)}{(0.348)(0.441)}$

$K_p=66.7$

Therefore, the value of $K_p$ at this temperature is 66.7

4 0

2 years ago

A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec

balu736 [363]

Answer:

E.F= OH

M.F= $O_{2} H_{2}$

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

$\frac{34.00}{17.008}= 1.999$ round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula. ANSWER: M.F=2(OH)- $O_{2} H_{2}$

3 0

3 years ago