How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)
I really couldn't find the answer since molarity and volume for sodium carbonate are not given.
I will mark the correct answer with steps as best answer.

Answers

Answer 1

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

Answer 2

The amount of HCl required for counterbalancing 1.25 g of Na2CO3(Sodium Carbonate) would be:

- [tex]34 ml[/tex]

Given that,

Mass of Na2CO3 [tex]= 1.25 g[/tex]

To find the Moles of Na2CO3, we will find the molar mass of Na2CO3,

Molar Mass of or Na2CO3 [tex]= 106 g/mol[/tex]

So,

Moles of Na2CO3 [tex]= mass /molar mass[/tex]

[tex]= 1.25/106[/tex]

[tex]= 0.012 mol[/tex]

To determine the quantity of HCl required to display the reaction with 0.012 mol of Na2CO3

[tex]Na_{2} CO_{2} + 2HCl[/tex][tex]H_{2}CO_{3} + 2NaCl[/tex]

While balancing the equation, we know that [tex]0.012[/tex] × [tex]2 = 0.024 mole of HCl[/tex] is necessary to process the reaction.

Now,

As we know,

HCl moles [tex]= 0.024[/tex]

Molarity of HCl [tex]= 0.715 M[/tex]

∵ Quantity of HCl required = Moles/Molarity

[tex]= 0.024 / 0.715[/tex]

[tex]= 0.034 l[/tex] [tex]or 34ml[/tex]

Thus, 34 ml is the correct answer.

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[tex]NH_3(g)+O_2(g)\rightarrow N_2O(g)+H_2O(l)+342kJ[/tex]

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Answers

Answer:

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Answers

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Answers

2.1 L

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Solve for V2 —> V2 = P1V1/P2
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The volume of the balloon is 2.1L.

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P1V1 = P2V2

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P1 = 3.5atm

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V2 = ?

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V2 = P1V1/P2

V2 = (3.5atm x 4.2 L) / (7.0atm)

     = 2.1L

Hence, the volume of the balloon is 2.1L.

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Answers

Answer:

1. Physical Weathering

2. Option B

Explanation:

1. Physical weathering

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2. Option B

3D can be used to produce things that are required on the Mars and also the locally available raw material can be used for printing thereby saving the cost.

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Answers

Answer:

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Answer:

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