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Vesnalui [34]
3 years ago
8

Please write a balanced equation for this reaction!! WILL GIVE BRAINLIEST!! 50 points!!

Chemistry
1 answer:
Alex73 [517]3 years ago
8 0

Answer:

Cu+2AgNO3--->2Ag+Cu(NO3)2

Explanation:

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Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One indu
trapecia [35]
3.8 mole Cl2
Explanation:
3.8molesC6H5Cl x 1moleCl2/ 1mole C6H5Cl
7 0
2 years ago
What is the atom inventory for the following equation after it is properly balanced? ____NaOH + ____CuCl2 Imported Asset ____NaC
faltersainse [42]

Answer:

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2;

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

Explanation:

NaOH + CuCL2 —> NaCl + Cu(OH)2

The balanced equation can be achieved by doing the following:

There are 2 oxygen and 2 hydrogen atom on the right side. This is balanced by putting 2 in front of NaOH as shown below:

2NaOH + CuCL2 —> NaCl + Cu(OH)2

This makes Na to be unbalanced. Now to balance Na, put 2 in front of NaCl as illustrated below

2NaOH + CuCL2 —> 2NaCl + Cu(OH)2

Now the equation is balanced.

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

4 0
3 years ago
Review this reaction:H2SO4 + NaOH ?.What are the product(s)?Na2SO4H2ONa2SO4 and H2ONaH and SO4OHH3SO5
muminat

i think its MIDDLE FINGERS UP IN THE SKY AND AT THESE AHOLE MODERATORS

6 0
3 years ago
Read 2 more answers
How many moles of HCl are present in 4.7 L of a 4.23 M HCl solution?
REY [17]

Hey There!

Volume = 4.7 L

Molarity = 4.23 M

Therefore:

n = M *V

n = 4.23 * 4.7

n = 19.881 moles of HCl

hope this helps!

6 0
3 years ago
How much (mL) of a 2.0 M sodium hydroxide solution would it take to neutralize 50 mL of a 6.0 M solution of hydrochloric acid?
Tom [10]

Answer:

150ml

Explanation:

For this question,

NaOH completely dissociates. It is a strong base

HCl also completely dissociates. It is a strong acid

So we have this equation

m1v1 = m2v2 ----> equation 1

M2 = 2m

V1= ??

M2 = 6m

V2 = 50m

When we input these into equation 1, we have:

2m x v1 = 6m x 50ml

V1 = 6m x 50ml/2

V1 = 300/2

V1 = 150ml

Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.

Thank you

8 0
2 years ago
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