Answer:
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Explanation:
Solubility of solute at 20°C = 32.2 g/100 grams of water
Solute soluble in 1 gram of water = ![\frac{32.2}{100}g=0.322 g](https://tex.z-dn.net/?f=%5Cfrac%7B32.2%7D%7B100%7Dg%3D0.322%20g)
Mass of solute in soluble in 56.0 grams of water:
![0.322\times 56.0=18.032 g](https://tex.z-dn.net/?f=0.322%5Ctimes%2056.0%3D18.032%20g)
Solubility of solute at 30°C = 70.2g/100 grams of water
Solute soluble in 1 gram of water = ![\frac{70.2}{100}g=0.702 g](https://tex.z-dn.net/?f=%5Cfrac%7B70.2%7D%7B100%7Dg%3D0.702%20g)
Mass of solute in soluble in 56.0 grams of water:
![0.702 \times 56.0=39.312 g](https://tex.z-dn.net/?f=0.702%20%5Ctimes%2056.0%3D39.312%20g)
If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C
Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g
Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g
Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:
39.312 g - 18.032 g = 21.28 g
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Amount of CH4 is excess, so no need to worry about it
<span>but the limiting factor is the Oxygen </span>
<span>according to stranded equation, </span>
<span>CH4 + 2 O2 --> CO2 + 2 H2O ΔH = -889 kJ/mol </span>
<span>just by taking proportions </span>
<span>(-889 kJ/mol) / 2 x 0.8 mol = - 355.6 kJ </span>
<span>so i think the answer is (a)</span>
Answer:
5.32 × 1017 nanometers
Explanation:
5.32 x (10^10) centimeters =
5.32 × 1017 nanometers
Answer:
covalent bonds
Explanation:
A chemical bond is a lasting attraction between atoms that enables the formation of chemical compounds and may result from the electrostatic force of attraction between atoms with opposite charges, or through the sharing of electrons as in the covalent bonds.