<span> Atoms combine as the electrons from each atom are attracted to the nuclei of the atoms. This results in bonds ranging from 100% covalent to bonds with high ionic character. The combination of atoms to form compounds occurs when the compounds being formed are at lower energy than the original atoms.</span>
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
625.46 °C
Explanation:
We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 19 °C
T(K) = 19 °C + 273
T(K) = 292 K
Next, we shall determine the Final temperature. This can be obtained as follow:
Initial volume (V₁) = 3.25 L
Initial temperature (T₁) = 292 K
Final volume (V₂) = 10 L
Final temperature (T₂) =?
V₁/T₁ = V₂/T₂
3.25 / 292 = 10 / T₂
Cross multiply
3.25 × T₂ = 292 × 10
3.25 × T₂ = 2920
Divide both side by 3.25
T₂ = 2920 / 3.25
T₂ = 898.46 K
Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 898.46 K
T(°C) = 898.46 – 273
T(°C) = 625.46 °C
Therefore the final temperature of the gas is 625.46 °C
Answer:
Copper ions are reduced into copper atoms.
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Explanation:
During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.
At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)
Thus the half equation is as follows:
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Answer:
V = 6.17 L
Explanation:
Given data:
Volume = ?
Number of moles = 0.382 mol
Pressure = 1.50 atm
Temperature = 295 k
R = 0.0821 L. atm. /mol. k
Solution:
According to ideal gas equation:
PV= nRT
V = nRT/P
V = 0.382 mol × 0.0821 L. atm. /mol. k ×295 k / 1.50 atm
V = 9.252 L. atm. / 1.50 atm
V = 6.17 L
