Answer:
<h2>9.03 × 10²³ molecules</h2>
Explanation:
The number of molecules can be found by using the formula
N = n × L
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 1.5 × 6.02 × 10²³
We have the final answer as
<h3>9.03 × 10²³ molecules</h3>
Hope this helps you
Answer:
0.7692 M ≅ 0.77 M.
Explanation:
- It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.
<em>(MV) before dilution = (MV) after dilution.</em>
M before dilution = 2.0 M, V before dilution = 0.25 L.
M after dilution = ??? M, V after dilution = 0.25 L + 0.40 L = 0.65 L.
∴<em> M after dilution = (MV) before dilution/(V) after dilution</em> = (2.0 M)(0.25 L)/(0.65 L) =<em> 0.7692 M ≅ 0.77 M.</em>
Suspension. The particles are big enough for the eye to see, and will separate if left sitting.
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 = 
Mass of solute in soluble in 56.0 grams of water:

Solubility of solute at 30°C = 70.2g/100 grams of water
Solute soluble in 1 gram of water = 
Mass of solute in soluble in 56.0 grams of water:

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.
False... Honey is greater in velosity then water