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ladessa [460]
3 years ago
14

A coefficient of "1" is understood. choose option "blank" for the correct answer if the coefficient is "1". co2 + h2o → c6h12o6

+ o2
Chemistry
2 answers:
tiny-mole [99]3 years ago
6 0
6CO2+6H2O=C6H12O6+6O2
Rashid [163]3 years ago
3 0

Answer : The balanced chemical reaction will be,

6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given unbalanced chemical reaction is,

CO_2+H_2O\rightarrow C_6H_{12}O_6+O_2

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of carbon, hydrogen and oxygen are not balanced.

In order to balanced the chemical reaction, the coefficient 6 is put before the CO_2,H_2O\text{ and }O_2.

Thus, the balanced chemical reaction will be,

6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2

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Explanation:

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IF YOU CAN HELP ME THAT WOULD BE GREAT
Likurg_2 [28]
Answer:
I. Changing the pressure:
Increasing the pressure: the amount of H₂S(g) will increase.
Decreasing the pressure: the amount of H₂S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of H₂S(g) will decrease.
Decreasing the temperature: the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
Increasing the H₂ concentration: the amount of H₂S(g) will increase.
Decreasing the H₂ concentration: the amount of H₂S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of H₂S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of H₂S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
H₂ is a part of the products.
Increasing the H₂ concentration:
H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.
Decreasing the H₂ concentration:
H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.
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Answer:

\boxed{\text{(a) 14 L; (b) 711 kg}}

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\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

\text{Kilograms of NaF}= \text{322 kg F}^{-} \times \dfrac{\text{41.99 kg NaF}}{\text{19.00 kg F}^{-}} = \text{711 kg NaF}\\\\\text{It will take } \boxed{\textbf{711 kg}} \text{ of NaF to treat the reservoir}

7 0
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