Answer:
Separating Sand and Salt
Probably the easiest method to separate the two substances is to dissolve salt in water, pour the liquid away from the sand, and then evaporate the water to recover the salt.
Answer:
a) 0.714g of bicarbonate of soda are required.
b) 0.221g of Al(OH)₃ are required
Explanation:
The reactions of HCl with bicarbonate of soda and aluminium hydroxide are:
HCl + NaHCO₃ → H₂O + NaCl + CO₂
3 HCl + Al(OH)₃ → 3H₂O + AlCl₃
The moles of HCl that we need neutralize are:
50mL = 0.050L * (0.17mol / L) = 0.0085 moles HCl
To solve these problem we need to find the moles of the antacid using the chemical reaction and its mass using its molar mass;
<em>a) </em><em>Moles NaHCO₃ = Moles HCl = 0.0085 moles </em>
The mass is -Molar mass NaHCO₃: -84g/mol-
0.0085 moles * (84g / mol) = 0.714g of bicarbonate of soda are required
b) 0.0085 moles HCl * (1mol Al(OH)₃ / 3mol HCl) = 2.83x10⁻³ moles Al(OH)₃
The mass is -Molar mass: 78g/mol-:
2.83x10⁻³ moles Al(OH)₃ * (78g/mol) =
<h3>0.221g of Al(OH)₃ are required</h3>
4.8 g/cm3 with sig figs since it's mass/volume you divide 76 grams by 16 cm3
Can you post the question
Answer:
Following are the responses to the given choices:
Explanation:
- The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
- 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
- Distilled H2 implies hemolytic distillation.
- Glucose is indicated by crenation at 8.69 percent (m/v).
- 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
