Nature of the solute and solvent - The amount of solute that dissolves depends on what type of solute it is. While only 1 gram of lead (II) chloride can be dissolved in 100 grams of water at room temperature, 200 grams of zinc chloride can be dissolved. This means that a greater amount of zinc chloride can be dissolved in the same amount of water than lead II chloride.
Temperature - Increasing the temperature not only increases the amount of solute that will dissolve but also increases the rate at which the solute will dissolve. For gases, the reverse is true. An increase in temperature decreases both solubility and rate of solution.
Pressure - Changes in pressure have practically no effect on solubility. For gaseous solutes, an increase in pressure increases solubility and a decrease in pressure decreases solubility. Example: When the cap on a bottle of soda pop is removed, pressure is released, and the gaseous solute bubbles out of solution. This escape of a gas from solution is called effervescence.
Stirring - Stirring brings fresh portions of the solvent in contact with the solute. Stirring, therefore, allows the solute to dissolve faster.
Hope this helps!! (If not I'm sorry!)
Weight is properly defined as how much gravity acts upon an object.
I hope this helps!
Answer:
(1) Chloroplast
Explanation:
Cells of living organisms are made up of certain function-specific structures called ORGANELLES. Some organelles are present in plant cells and absent in animal cells and vice versa. In a plant cell, one notable organelle that allows it perform the photosynthetic process is the CHLOROPLAST.
However, the chloroplast is predominantly found in the LEAF part of a plant. This is because leaf cells are the site of photosynthesis. Hence, according to this question, Joe would be able to tell whether the plant cell was from the leaf or the root by looking for CHLOROPLAST as a differentiating factor in each cell.
The answer that i found is 4.650
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
