C & B are switched so I'm not sure if that was a typo or not, but the answer is concentration!
.sand and water
.iron fillings in a liquid
.cereal in milk
Elements are substances that are made up of the same atoms which are capable of taking part in a chemical reaction.
There are different types of elements which are represented by symbols gotten from the first letter or the first and any other letter in the name of the element.
Examples of elements include:
When two or more of these elements combine together through a chemical bond, it leads to the formation of compounds.
Example of a compound includes:
- NaCl: The element sodium combine, through electrochemical bonding, with another element chlorine to form the compound sodium chloride.
Learn more here:
brainly.com/question/17571315
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion
1) Content of Ca (2+) ions
Calcium chloride = CaCl2
Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)
=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)
Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution
M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2
=> 0.0825 mol Ca(2+)
2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)
formula of phospahte ion: PO4 (3-)
molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2
Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)
=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)
3) Content of Mg(2+) ions
Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)
Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)
number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution
n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2
ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)
4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)
2PO4(3-) + 3Mg(2+) = Mg3(PO4)2
=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)
=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)
5) Total number of moles of PO4(3-)
0.055 mol + 0.16 mol = 0.215 mol
6) Sodium phosphate
Sodium phosphate = Na3(PO4)
Na3PO4 ---> 3Na(+) + PO4(3-)
=> 1 mol Na3PO4 : 1 mol PO4(3-)
=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4
mass in grams = number of moles * molar mass
molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol
=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g
Answer: 35.26 g of sodium phosphate
</span>
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.