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Yuliya22 [10]
3 years ago
7

What parts are present in every chemical equation?

Chemistry
1 answer:
WITCHER [35]3 years ago
5 0
The reactants, products, coefficients, subscripts. ( I forgot the rest lol)
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Which Group is in the leftmost column on the periodic table?
svp [43]
Alkali metals. Hope this helps
4 0
3 years ago
Read 2 more answers
PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
anastassius [24]

Answer:

I believe the first one is correct but I think the second one would be the blood like substance. not a 100% sure sorry

6 0
3 years ago
I want to know which ones are molecular equation, complete ionic equation and net ionic equation
NNADVOKAT [17]

Answer:

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

8 0
3 years ago
How many moles would be in 85.OmL of 0.750M KOH?
Harlamova29_29 [7]

Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.

Explanation:

Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L

Molarity = 0.750 M

It is known that molarity is the number of moles of solute present in liter of a solution.

Therefore, moles present in given solution are calculated as follows.

Molarity = \frac{moles}{Volume (in L)}\\0.750 M = \frac{moles}{0.085 L}\\moles = 0.0637 mol

Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.

7 0
3 years ago
In which of the following is the solution concentration expressed in terms of molarity?
puteri [66]
Answer:
D. <span>10 mol of solute/ 1L of solution

Explanation:
Molarity is used to express the concentration of the solution. It is expressed as the number of moles of solute per liter of solution

Molarity = moles of solute / liters of solution
This means that molarity has the units moles/liter

The only option satisfying this is the last one.

Hope this helps :)</span>
8 0
3 years ago
Read 2 more answers
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