Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
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The solution has a concentration 20 mgr in each mL of the final solution.
To solve this problem, we need to know about concentration. The concentration formula can be defined as how much the mass per unit volume is. It can be written as
M = m/V
where M is concentration, m is mass of solute, V is the total volume of solution.
From the text we know that :
m = 10g
vsolvent = 45mL
vsolute = 5 mL
find the total volume (V)
V = vsolvent + vsolute
V = 45 + 5
V = 50mL
Then, find the concentration
M = m/V
M = 10gr / 50 mL
M = 1000 mgr / 50mL
M = 20 mgr / mL
Hence, the solution has a concentration 20 mgr in each mL of the final solution.
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The molality of a solution containing 3. 0 moles of NaCl and 100. 0 moles of water is 30 mol/kg.
The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality. Mole per kilogram of solvent is the SI unit for molality.
Given:
3.0 moles of NaCl in 100 moles of water.
To find:
The molality of the solution
The moles of solute (NaCl) = 3.0 moles
The mass of solvent (water) = 100 moles (0.1 kg/mol)
Molality of a solution = Number of Moles of solute/ Mass of solvent(kg)
= 3.0 moles/0.1 kg/mol
= 30 mol/kg
To know more about Molality refer:
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Answer:
1- Using buses rather than opting for cars.
2- Regularly servicing your cars.
Explanation: