The chemical equation can be expressed as:
2H2 + O2 = 2H2O
Given the amounts of the reactants, we need to identify the limiting reactant before calculating the amount of product generated.
4.0 g H2 ( 1 mol / 2.02 g ) = 1.98 mol H2
5.0 g O2 ( 1 mol / 32 g ) = 0.1563 mol O2
The limiting reactant is O2, as it will be fully consumed in the reaction.
0.1563 mol O2 ( 2 mol H2O / 1 mol O2 ) ( 18.02 g / mol ) = 5.6 g H2O will be produced
Given parameters:
Mass of sucrose = 5g
Density of sucrose = 1.12g/mL
Percentage of sucrose per liter of cane juice = 12%
Unknown:
Volume of cane juice required =?
We need to understand the relationship between volume and density. Density represents mass per unit volume.
Mathematically;
Density =
Now, calculate the volume of sucrose;
1.12g/mL =
Volume =
= 4.46mL = 4.46 x 10⁻³L since 1000mL = 1L
Since 12% of one liter of cane juice is sucrose,
12% of x liter of cane juice = 4.46 x 10⁻³L
Volume of cane juice = 4.46 x 10⁻³ x
= 0.037L
Volume of cane juice needed is 0.037L
The issue with your setup is that the surface where you placed the drop is contaminated or dirty, preventing it from being wetted by the solution. To fix this, thoroughly clean the surface before applying the drop.
Answer:
By reducing the height of the center of gravity of the object in relation to its center of buoyancy
Explanation:
In the field of hydrostatics, for a floating object, the state of equilibrium corresponds to either a peak or a trough in potential energy. Stability in equilibrium occurs when the potential energy is minimized. Achieving a lower position of the center of gravity of the floating object compared to its center of buoyancy creates a stable equilibrium arrangement.