Answer:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Explanation:
Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).
The equation can be written as follow:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
The above equation can be balance as illustrated below:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O
There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O
There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Now, the equation is balanced.
Answer:
basic solution.
Explanation:
has a pH of 12.
it turns clear indicator pink.
it contains hydroxide ions which conduct electricity
Aueyeowosy yo this is my answer
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer: If a substance has a boiling point of
then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.
Explanation:
The temperature at which vapor pressure of a liquid substance becomes equal to the atmospheric pressure is called boiling point of substance.
At the boiling point, liquid phase and vapor phase remains in equilibrium.
This means that as liquid phase changes into vapor phase and also vapor phase changes into liquid phase at the boiling point.
Thus, we can conclude that if a substance has a boiling point of
then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.