<u>Answer:</u> The density of NaCl solution is 3.930 g/mL
<u>Explanation:</u>
We are given:
Mass of cylinder,
= 21.577 g
Mass of NaCl and cylinder combined, M = 39.664 g
Mass of NaCl,
= 
To calculate density of a substance, we use the equation:

We are given:
Mass of NaCl = 18.087 g
Volume of NaCl solution = 4.602 mL
Putting values in above equation, we get:

Hence, the density of NaCl solution is 3.930 g/mL
Answer:
It is a beta decay equation unknown
Explanation:
none
Answer:
pineal body is a vastigious structure which signifies same origin of chimpanzee and modern man
Explanation:
Answer:
0.808 M
Explanation:
Using Raoult's Law

where:
= vapor pressure of sea water( solution) = 23.09 mmHg
= vapor pressure of pure water (solute) = 23.76 mmHg
= mole fraction of water
∴






------ equation (1)
------ equation (2)
where;
number of moles of sea water
number of moles of pure water
equating above equation 1 and 2; we have :



NOW, Molarity = 



As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have 
The number in standard form is 0.480