<span>Ores are naturally occurring rocks that contain metal or metal compounds in sufficient amounts to make it worthwhile extracting them. For example, iron ore is used to make iron and steel.</span>
Answer:
See the answers below
Explanation:
1) 100. mL of solution containing 19.5 g of NaCl (3.3M)
2) 100. mL of 3.00 M NaCl solution (3 M)
3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)
4) Number 1 and 5 have the same concentration (1.5M)
MW of NaCl = 23 + 36 = 59 g
For number 3
59 g ------------------- 1 mol
19,5 g ----------------- x
x = 19.5 x 1/59 = 0.33 mol
Molarity (M) = 0.33 mol/0.150 l = 2.2 M
For number 4,
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For number 5
Molarity (M) = 0.450/0.3 = 1.5 M
The scientists would do biological studies of how the protein breakdown and combines with the muscles the engineers with then create a delivery system to get the protein to the muscle quicker and more effectively
Complete question is;
Chemical reactivity of alkali metals increases down the group while reactivity of halogens decreases down the group. Give reasons
Answer:
Explained below
Explanation:
Alkali metals exhibit reactivity due to their electropositivity. Now, for alkalis, their electro-positivity increases down their group. Since their reactivity increases with increase in electropositivity, it means their reactivity also increases down the group.
Whereas, the reactivity of halogens occurs as a result of their electronegativity. Now, electronegativity for halogens decreases down the group. Since their reactivity decreases with decrease in electronegativity, it means that their reactivity will also decrease down the group.
Answer:
0.6 moles of CaO will produced.
Explanation:
Given data:
Mass of calcium = 23.9 g
Moles of CaO produced = ?
Solution:
Chemical equation:
2Ca + O₂ → 2CaO
Number of moles of calcium:
Number of moles = mass/ molar mass
Number of moles = 23.9 g / 40 g/mol
Number of moles = 0.6 mol
Now we will compare the moles of calcium and CaO.
Ca : CaO
2 : 2
0.6 : 0.6
0.6 moles of CaO will produced.
