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Korvikt [17]
3 years ago
12

How many moles are in 2.5L of 1.75 M Na2CO3

Chemistry
1 answer:
mixas84 [53]3 years ago
6 0

Answer: There are 4.375 moles in 2.5 L of 1.75 M Na_2CO_3

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}      

Molarity of solution = 1.75 M

Volume of solution = 2.5 L  

Putting values in equation , we get:

1.75M=\frac{\text{Moles of} Na_2CO_3}{2.5L}\\\\\text{Moles of }Na_2CO_3=1.75mol/L\times 2.5L=4.375mol

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A twenty-eight-liter volume of gas contains 11 g methane, 1.5-gram nitrogen and 16-gram carbon dioxide. Determine partial pressu
GREYUIT [131]

Based on Dalton's Law, for a mixture of gases, the total pressure is the sum of the partial pressure of each gas.

Partial pressure (p) of each gas is related to the total pressure (P) as follows:

p = X * P----------(1)

where X is the mole fraction of that gas

X = moles of a particular gas/total number of moles of all gases in the mixture--------------(2)

Step 1: Calculate the moles of each gas

Mass of methane, CH4 = 11 g

Mass of nitrogen, N2 = 1.4 g

Mass of carbon dioxide, CO2 = 16 g

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# moles of CO2 = 16/44 = 0.3636

Total moles = 0.6875+0.05+0.3636 = 1.1011

Step2: Calculate mole fractions of each gas

Based on equation (2)

X(CH4) = 0.6875/1.1011 = 0.6244

X(N2) = 0.05/1.1011 = 0.0454

X(CO2) = 0.3636/1.1011 = 0.3302

Step 3: Calculate the total pressure

Based on ideal gas equation:

PV = nRT

given that;

V = 28 L

n = total moles = 1.1011

R = gas constant = 0.0821 Latm/mol-K

Since temp T is not given, let us consider room temperature of 25 C = 25 + 273 = 298 K

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Step 3: Calculate partial pressures

Based on equation:

p(CH4) = 0.6244*0.962 atm = 0.601 atm

P(N2) = 0.0454*0.962 atm = 0.044 atm

P(CO2) = 0.3302*0.962 atm = 0.318 atm


4 0
3 years ago
A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva
Sergeu [11.5K]

Answer:

  • <u>0.1255 M</u>

Explanation:

<u>1) Data:</u>

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

<u>2) Chemical reaction:</u>

The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:

  • Acid + Base → Salt + Water

  • H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

<u>3) Balanced chemical equation:</u>

  • H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

<u>4) Stoichiometric mole ratio:</u>

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

  • 1 mole H₂SO₄ : 2 mole NaOH

<u>5) Calculations:</u>

a) Molarity formula: M = n / V (in liter)

   ⇒ n = M × V

b) Nunber of moles of acid:

  • nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

  • nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

  • nₐ = nb

  • 0.1053 × (17.88 / 1,000) =  Mb × (15.00 / 1,000)

  • Mb = 0.1053 × 17.88  / 15.00 = 0.1255 mole/liter = 0.1255 M
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