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KengaRu [80]
3 years ago
15

Consider calcium nitrate, Ca(NO3)2. Choose the best explanation for the subscript, 2, from the list provided There are two atoms

of oxygen. There are two atoms of nitrogen. There are two nitrogen atoms and five oxygen atoms. There are two nitrate ions.
Chemistry
2 answers:
xxMikexx [17]3 years ago
7 0

D is correct There are two nitrate ions.

Fill in the blank with the molar mass of calcium nitrate. A link to the periodic table is provided for you.

⇒ 164.10 g/mol

lutik1710 [3]3 years ago
5 0

Answer : The correct option is, There are two nitrate ions.

Explanation :

Calcium nitrate is formed by the combination of a metal cation and poly-atomic anion. On the dissociation, it will dissociates into one calcium ion and two nitrate ions.

Ca(NO_3)_2\rightarrow Ca^{2+}+2NO^-_3

Hence, the subscript, 2 indicates that there are two nitrate ions in calcium nitrate.

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How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq
Alexxx [7]

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

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what is the molarity of a solution made by dissolving 18.5 grams of NaCL into enough water to reach 450.0ML​
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Answer:

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Molarity of a solution, which refers to the molar concentration of that solution can be calculated thus;

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