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3 years ago

How many liters of hydrogen gas I needed to react with CS2 to produce 2.5 L of CH4 At STP

Chemistry

1 answer:

olchik [2.2K]3 years ago

8 0

Answer:

9.8 L

Explanation:

The reaction that takes place is:

4H₂(g) + CS₂(g) → CH₄(g)+ 2H₂S(g)

At STP, 1 mol of any gas occupies 22.4 L.

We calculate how many moles are there in 2.5 L of CH₄ at STP:

2.5 L ÷ 22.4 L/mol = 0.11 mol CH₄

Then we convert CH₄ moles into H₂ moles, using the stoichiometric coefficients of the reaction:

0.11 mol CH₄ * $\frac{4molH_2}{1molCH_4}$ = 0.44 mol H₂

Finally we calculate the volume that 0.44 moles of H₂ would occupy at STP:

0.44 mol * 22.4 L/mol = 9.8 L

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Suppose you have 1.00 l of an aqueous buffer containing 60.0 mmol acetic acid (pka = 4.76) and 40.0 mmol acetate. what volume of

butalik [34]

Required pH = 4.93
- OH⁻ from NaOH reacts with CH₃COOH giving CH₃COO⁻ and H₂O
- Let the volume of 3.5 M NaOH be x ml
Moles of NaOH = Moles of OH⁻ = Molarity * x ml = 3.5x mmol
- The reaction table for moles is as follows:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Initial 60 3.5x 40
Change -3.5x -3.5x +3.5x
Final (60-3.5x) 0 (40+3.5x)
- Substitute in Henderson equation and solve for x:
pH = pKa + log $\frac{[CH_{3}COO^-]}{[CH_3COOH]}$
4.93 = 4.76 + log $\frac{(40+3.5x)}{(60-3.5x)}$
0.17 = log $\frac{(40+3.5x)}{(60-3.5x)}$
$\frac{(40+3.5x)}{(60-3.5x)} = 1.479$
x = 5.62 ml NaOH required

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3 years ago

calculate the molality of commercial HCl solution which is 12.1Molarity, and has a density of 1.19g/mL, and is 37.2wt.%HCl.

shutvik [7]

Answer:

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4 years ago

During the titration of a weak acid and a strong base, you would expect the ph at the endpoint of titration to be ____.

riadik2000 [5.3K]

Answer is: b. more than 7.

The endpoint is the point at which the indicator changes colour in a colourimetric titration and that is point when titration must stop.

For example, basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).

Balanced chemical reaction of acetic acid and sodium hydroxide:

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).

Neutralization is is reaction in which an acid (in this example vinegar or acetic acid CH₃COOH) and a base react quantitatively with each other.

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3 years ago

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Water is a polar molecule, meaning it carries partial charges (Î´ or Î´â€“) on opposite sides of the molecule. for two formula u

Zielflug [23.3K]

The side of each water molecule with the oxygen atom uncovered will be marginally negative.
The side of each water molecule with the hydrogen atoms uncovered will be marginally positive.
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3 years ago

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What is the molar concentration (in moles/L) of a solution made with 3.744 g of Mg(NO3)2 dissolved in enough water to make 50.0

Annette [7]

Answer:

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is $0.5 \frac{moles}{L}$

Explanation:

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$

In this case:

Mg: 24.3 g/mole
N: 14 g/mole
O: 16 g/mole

So, the molar mass of Mg(NO₃)₂ is:

Mg(NO₃)₂= 24.3 g/mole + 2*(14 g/mole + 3*16 g/mole)= 148.3 g/mole

So, if you have 3.744 g of Mg(NO₃)₂, you can apply the following rule of three: if 148.3 grams of Mg(NO₃)₂ are present in 1 mole, 3.744 grams in how many moles are present?

$moles=\frac{3.744 grams*1mole}{148.3 grams}$

moles= 0.025

Then you have:

number of moles=0.025
volume= 50 mL= 0.05 L (being 1,000 mL= 1 L)

Replacing in the definition of molarity:

$molarity=\frac{0.025 moles}{0.05 L}$

you get:

$molarity=0.5 \frac{moles}{L}$

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is $0.5 \frac{moles}{L}$

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3 years ago