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Dimas [21]
3 years ago
13

Complete the following equations (note that the equations are not balanced). Use the activity series if necessary. An activity s

eries, listed as most reactive to least reactive: Upper L i, Upper K, Upper B a, Upper S r, Upper C a, Upper N a, Upper M g, Upper A l, Upper M n, Upper Z n, Upper C r, Upper F e, Upper C d, Upper C o, Upper N i, Upper S n, Upper P b, Upper H, Upper S b, Upper B i, Upper C u, Upper A g, Upper P d, Upper H g, Upper P t, Upper A u. PbCl2 MgSO4Upper C subscript 3 upper H subscript 8 plus upper O subscript 2 right arrow. ? Pb O2 MgCl PbMg ClSO4 PbSO4 MgCl2 Pb(SO4)2 MgCl.
SAT
2 answers:
balandron [24]3 years ago
5 0

Answer:

PbSO4 + MgCl2

Explanation:

got it right on edge

Anestetic [448]3 years ago
3 0

Answer:

<em>hey hope your day is amazing and i hope my answer to your question is helpful :)</em>

Explanation:

<em>***for the first question</em> on  this slide is C.) <em>PbSO₄ + MgCI₂</em>

***<em>for the second question</em> on this slide its going to be B.)<em> </em><em>MgCI₂</em>

<em>i hope you have a wonderful rest of your day :D !</em>

        \\\\<em>***~yurGurlAngieee~***////</em>

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a student combined 45.3 ml of a 0.549 m calcium nitrate ca(no₃)₂ solution with 75.05 ml of a 1.321 m ca(no₃)₂ solution. calculat
Maksim231197 [3]

The concentration of the final solution of Ca(NO₃)₂ made by combining 45.3 mL of a 0.549 M Ca(NO₃)₂ solution with 75.05 mL of a 1.321 M Ca(NO₃)₂ solution is 1.03 M

We'll begin by calculating the number of mole of Ca(NO₃)₂ in each solution. This can be obtained as follow:

<h3>For solution 1:</h3>

Volume = 45.3 mL = 45.3 / 1000 = 0.0453 L

Molarity = 0.549 M

<h3>Mole of Ca(NO₃)₂ =?</h3>

Mole = Molarity x Volume

Mole of Ca(NO₃)₂ = 0.549 × 0.0453

<h3>Mole of Ca(NO₃)₂ = 0.0249 mole</h3>

<h3>For solution 2:</h3>

Volume = 75.05 mL = 75.05 / 1000 = 0.07505 L

Molarity = 1.321 M

<h3>Mole of Ca(NO₃)₂ =?</h3>

Mole = Molarity x Volume

Mole of Ca(NO₃)₂ = 1.321 × 0.07505

<h3>Mole of Ca(NO₃)₂ = 0.0991 mole</h3>

  • Next, we shall determine the total mole of Ca(NO₃)₂ in the final solution. This can be obtained as follow:

Mole of Ca(NO₃)₂ in solution 1 = 0.0249 mole

Mole of Ca(NO₃)₂ in solution 2 = 0.0991 mole

Total mole = 0.0249 + 0.0991

<h3>Total mole = 0.124 mole</h3>

  • Next, we shall determine the total volume of the final solution.

Volume of solution 1 = 0.0453 L

Volume of solution 2 = 0.07505 L

Total Volume = 0.0453 + 0.07505

<h3>Total Volume = 0.12035 L</h3>

  • Finally, we shall determine the concentration of the final solution. This can be obtained as follow:

Total Volume = 0.12035 L

Total mole = 0.124 mole

<h3>Concentration =?</h3>

Concentration = mole / Volume

Concentration = 0.124 / 0.12035

<h3>Concentration = 1.03 M</h3>

Therefore, the concentration of the final solution of Ca(NO₃)₂ is 1.03 M

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<h3>What is the industry incubation stage?</h3>

The industry incubation stage can be defined as the period between when an industry experiences a technological breakthrough of its first product and the first instance of its commercialization.

In Business management, pre-entrants to the industry incubation stage typically make use of strategic alliances and acquisitions by developing new core technologies to create a portfolio of various sourcing modes.

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In Computer science, some examples of a computing device include the following:

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