First solve the mass of 37.9 L seawater
M = 37.9 L ( 1000 ml / 1L ) ( 1.03 g / ml)
M = 39037 g
Then convert it to lb
M = 39037 g ( 1 lb / 453.592 g)
M = 86.06 lb
<span>Total weight = 86.06 lb + 59.5 lb = 145.56 lb so it can
support the aquarium</span>
They are arranged in shells
There are 17.97 moles of calcium chloride would react with 5. 99 moles of aluminum oxide .
The balanced chemical equation between reaction between calcium chloride and aluminum oxide is given as,
→ 
The molar ratio of above reaction is 3:1
It means 3 moles of calcium chloride is require to react one mole of aluminum oxide.
The number of moles of calcium chloride requires to react with 5. 99 moles of aluminum oxide = 3 × 5. 99 = 17.97 moles
The equation in which number of atoms of elements in reactant side is equal to the number of atoms of elements in product side is called balanced chemical equation .
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We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:
osmotic pressure = CRT
<span>C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L</span>
<span>moles of particles = C*V = 0.3189*0.250 =0.0797 mol </span>
<span>0.0797 = moles of sucrose + 2*moles of salt </span>
<span>x + 2y = 0.0797 </span>
<span>and </span>
<span>x(MMsucrose) + y(MMNaCl) = 10.2</span>
<span>342x + 58.5y = 10.2
</span>
<span>solve for x and y
</span>
<span>x = 0.0252 mol sucrose</span>
<span>y = 0.0273 mol NaCl
</span>
<span>mass Sucrose = 0.0252(342) = 8.6184 g </span>
<span>mass NaCl = 0.0273(58.5) = 1.5971 g </span>
<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>
Some chemical changes can be reactive through another chemical change?