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Cloud [144]
3 years ago
7

How much 3.0 M NaOH is needed to neutralize 30. mL of 0.75 M H2SO4?

Chemistry
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

15.0 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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Draw a structural isomer of ethylcyclohexane that also contains a 6-carbon ring
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1,2-methylcyclohexane, 1,3-methylcyclohexane, 1,4-methylcyclohexane
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3 years ago
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Please help ASAP please and thanks!! I'll also mark you!
skelet666 [1.2K]
Substitution Reactions are those reactions in which one nucleophile replaces another nucleophile present on a substrate. These reactions can take place via two different mechanism i.e SN¹ or SN². In SN¹ substitution reactions the leaving group leaves first forming a carbocation and nucleophile attacks carbocation in the second step. While in SN² reactions the addition of Nucleophile and leaving of leaving group take place simultaneously.
 
Example:
                        OH⁻  +  CH₃-Br     →     CH₃-OH  +  Br⁻

In above reaction,

                   OH⁻  = Incoming Nucleophile

                   CH₃-Br  =  Substrate

                   CH₃-OH  =  Product

                   Br⁻  =  Leaving group

Organic reactions are typically slower than ionic reactions because in organic compounds the covalent bonds are first broken, this breaking of bonds is a slower step, while, in ionic compounds no bond breakage is required as it consists of ions, so only bond formation takes place which is a quicker and fast step.
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3 years ago
Calculating Density Warm Up
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Explanation:

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6 0
3 years ago
Select all that apply.
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I think it is A and C!
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2 years ago
What is the mass of 3.20x10^23 formula units of iron (III) oxide (Fe2O3)?
yaroslaw [1]

The mass of iron (III) oxide (Fe2O3) : 85.12 g

<h3>Further explanation</h3>

Given

3.20x10²³ formula units

Required

The mass

Solution

1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol of Fe₂O₃ :

\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532

mass of Fe₂O₃ (MW=160 g/mol)

\tt mass=mol\times MW=0.532\times 160=85.12~g

4 0
2 years ago
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