Acid-Base Reactions Custom Learning Goal: To learn to classify acids and bases and to predict the products of neutralization rea
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Answer:
<u>1.364 M</u>.
Explanation:
Molarity formula: M= n/v, where n is moles of solute, and v is liters of solution.
Now, we need to convert the grams of nickel to moles and the volume of water to liters.
125mL/1000= 0.125 L.
To convert nickel grams to moles, we need to take a look at it's chemical formula, which is:
Now we count how many molecules of each element we have:
Ni= 1
N= 2
O= 6
Calculate the weight (g) of each element (the values of g/mol can be found on the periodic table and they may vary slightly between one table and the other):
Ni: (1) (58.6934)= 58.6934
N= (2) (14.007)= 28.014
O= (6) (15.999)= 95.994
Sum all the values to obtain the total weight of 1 mole of this compound:
58.6934+28.014+95.994= 129(g/mole)
Now that we know the that 129 grams equal 1 mole of nickel(II) nitrate, we can convert the 22.0 g to moles:
129g ------- 1 mole
22.0g ----- x
x= (22*1)/129= 0.1705 moles.
Now, we have all the values needed to calculate the molarity of this solution. All we have to do is substitute the values in the formula:
M= (0.1705 moles) / (0.125 L)= <u>1.364 M</u>.
If every oxygen ion is combined with an aluminium ion has a charge of -2,the charge of each aluminum ion would be -3.
Uncharged Aluminum atom must need to lose it's electrons,in order to form the bond with oxygen which has vacant orbitals
ion
atom that has a positive or negative charge because it lost or gained one or more electrons
chemical bond
the attractive force that holds atoms or ions together
ionic bond
a chemical bond in which one atom loses an electron and the other atom gains electrons to form ions
chemical formula
a combination of chemical symbols and numbers to represent a substance
covalent bond
bond formed by the sharing of electrons between atoms
Explanation:
because there are 4 Iodines on the left, we'll put. 4 in front of NaI to balance it. This would result in 4 Na on the left, so we'll put a 2 in front of Sodium Sulfate to balance the right side. Now we have 4 Na and I on both side, as well as 2 Sulfate on both sides. Pb is already balanced. The equation is now complete.
