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sammy [17]
3 years ago
15

Enter a balanced equation for the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II

) chloride and aqueous lithium nitrite.
Chemistry
1 answer:
Dominik [7]3 years ago
4 0

Answer:

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

Explanation:

Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.

Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

Now, we have to balance Li by multiplying LiNO₂ by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

The equation is now balanced.

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An insect needs mud wallows to breed; that insect pollinates several flowers in that ecosystem.

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QUESTION 3 (a) Ammonium sulphate, (NH),50, is a soluble salt and it is used in agriculture as fertiliser. 5 g of ammonium sulpha
nataly862011 [7]

Answer:

The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)

The number of moles = 5 g / 132.14 g/mol = 0.038 mol

The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23

the number of positive ions present in the ammonium sulphate solution:

2 positive ions for every 1 molecule of (NH₄)₂SO₄

so 2 x 2.29x10^23 = 4.58x10^23

the number of negative ions present in the ammonium sulphate solution

1 negative ion for every 1 molecule of (NH₄)₂SO₄

so 1 x 2.29x10^23 = 2.29x10^23

the total number of ions present in the ammonium sulphate solution​

4.58x10^23 + 2.29x10^23 = 6.87x10^23

4 0
3 years ago
Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
katrin2010 [14]

Answer:

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

The ratio here for both calcium chloride and magnesium nitrate is 3:2

The number of moles of each compound is equal to

0.054 * 1.4 = 0.0756\\0.093* 1.4 = 0.1302

Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

0.0756 mole of CaCl2 is equal to 0.05\\ Na3PO4

0.1302 mole of CaCl2 is equal to 0.0868 Na3PO4

Converting moles of sodium phosphate to grams of sodium phosphate we get

(0.05 +0.0868) * 163.94 g/mol

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

8 0
3 years ago
If the half-life of a radioisotope is 10,000 years, the amount remaining after 20,000 years is
Cloud [144]

Answer:

Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.

Explanation:

The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:

Time_________________ Amount

t=0_____________________x

t=10,000 years____________x/2

t=20,000 years___________x/4

During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.

6 0
3 years ago
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