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Alex17521 [72]
2 years ago
12

I need help with this question 20 points and brainiest if ur right

Chemistry
2 answers:
Sonbull [250]2 years ago
8 0

Hey How are you?

Explanation:

Nutka1998 [239]2 years ago
3 0

Answer:

The volume is 8 cm³

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Which of the following is a balanced chemical equation?
True [87]

Answer:

equation number 3 is balanced.

hope it helps ☺️!

8 0
3 years ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
What is the name of this compound
lys-0071 [83]

Answer:

horizontal

Explanation:

5 0
3 years ago
Would it be C..? <br> More NO2 and SO2 would form..help..please and thank you
Scorpion4ik [409]

Answer:

C. More NO2 and SO2 will form

Explanation:

Le Chatelier's Principle : It  predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc

It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.

In given Reaction SO3 is introduced(increased) .

So equilibrium will shift in the direction where SO3 should be consumed(decreased)

Hence the equilibrium will go in backward direction , i.e

NO + SO_{3} \rightarrow NO_{2} + SO_{2}

So more and more Of NO2 and SO2 will form

7 0
3 years ago
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

7 0
3 years ago
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