Answer:
2.69 x 10²³ atoms
Explanation:
Given parameters:
Volume of sea water = 2.06mL
number of moles of gold = 0.446moles
unknown:
Number of atoms in the sea water
Solution:
The mole is a convenient unit of measuring quantities of particles. In a mole we have 6.02 x 10²³particles.
The mole can be defined as the amount of a substance contained in Avogadro's number of particles which is 6.02 x 10²³particles.
Here, the particles we want to determine is the number of atoms:
Number of atoms is given as:
Number of gold atoms = number of moles of gold x 6.02 x 10²³
Number of gold atoms = 0.446 x 6.02 x 10²³ = 2.69 x 10²³ atoms
Molar mass NaOH =23+16+1=40 g/mol
<span> 0.100 M= 0.100 mol/L
</span>500 ml=0.500 L
0.500L*0.100 mol/L=0.0500 mol NaOH we need to prepare 500 ml solution
0.0500 mol NaOH*40g/1mol=2 g NaOH we need to prepare 500 ml solution
we need 2 g NaOH, dissolve it in small amount of water, and dilute it with water up to 500 mL
<span>Each liquid boils only at a certain temperature, which is called its </span><span>Boiling Point
Water for example has a boiling point of 100 degrees Celsius.</span>
69.9%
Explanation:
To find the mass percentage of iron in the compound in Fe₂O₃, we would go ahead to express the given molar mass of the iron to that of the compound.
Mass percentage = 
x 100
Molar mass of Fe = 55.85g/mol
Molar mass of O = 16g/mol
Molar mass of Fe₂O₃ = 2(55.85) + 3(16) = 159.7g/mol
Mass percentage = 
= 69.94% = 69.9%
learn more:
Mass percentage brainly.com/question/8170905
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Answer:
Solution A: 3.21% (m/v) NaCl ⇒ CRENATION
Solution B: 1.65% (m/v) glucose ⇒ HEMOLYSIS
Solution C: distilled H₂O ⇒ HEMOLYSIS
Solution D: 6.97% (m/v) glucose ⇒ CRENATION
Solution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl ⇒ CRENATION
Explanation:
Isotonic solution = 0.9% (m/v) NaCl or 5.0% (m/v) glucose
<u>Crenation</u> will occur if the solution has a concentration higher than 0.9% (m/v) NaCl or 5.0% (m/v) glucose (hypertonic solution). This will occur in the following solutions:
Solution A: 3.21% (m/v) NaCl > 0.9% (m/v) NaCl
Solution D: 6.97% (m/v) glucose > 5.0% (m/v) glucose
Solution E: 5.0% (m/v) glucose <u>and</u> 0.9%(m/v) NaCl (the addition of the 2 components exceeds the osmotic pressure permitted).
<u>Hemolysis</u> will occur if the solution has a concentration lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose (hypotonic solution). This is the case of:
Solution B: 1.65% (m/v) glucose < 5.0% (m/v) glucose
Solution C: distilled H₂O = 0% glucose/0% NaCl < 0.9% (m/v) NaCl or 5.0% (m/v) glucose
