When two atoms with equal electronegativity bond together, they form nonpolar covalent bonds.
Your second statement mirrors the first; the second statement likely reads, "Bonds between two atoms with unequal electronegativity are termed polar covalent bonds."
Convert 55.0g Ca(OH)2 to moles.
The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.
To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.
If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
Answer:
Please review the following responses
Explanation:
1) A solution of 100. mL contains 19.5 g of NaCl (3.3M)
2) 100. mL of NaCl solution at 3.00 M (3 M)
3) A solution of 150. mL holds 19.5 g of NaCl (2.2 M)
4) The concentrations of beakers 1 and 5 are identical (1.5M)
Molar mass of NaCl = 23 + 36 = 59 g
For beaker 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 beaker number 4:
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For beaker number 5:
Molarity (M) = 0.450/0.3 = 1.5 M
Clarification:
To obtain the specific element, you should multiply the grams provided by the ratio of grams of that particular element within its complete compound.
Since the query did not indicate the amount of NO2 produced, we can consider its mass to be negligible, thus assigning 1 mole to Nitrogen.
<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.
Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>
