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A 0.1873 g sample of a pure, solid acid, H2X (a diprotic acid) was dissolved in water and titrated with 0.1052 M NaOH solution.

puteri [66]

Answer:

We need 41.8 mL of NaOH

Explanation:

Step 1: Data given

Mass of H2X = 0.1873 grams

Molarity of NaOH solution = 0.1052 M

Molar mass of H2X = 85.00 g/mol

Step 2: The balanced equation

H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)

Step 3: Calculate moles of H2X

Moles H2X = mass H2X / Molar mass H2X

Moles H2X = 0.1873 grams / 85.00 g/mol

Moles H2X = 0.0022 moles

Step 4: Calculate moles of NaOH

For 1 mol H2X we need 2 moles NaOH to produce 1 mole of Na2X and 2 moles of H2O

For 0.0022 moles of H2X we need 0.0044 moles of NaOH

Step 5: Calculate volume of NaOH

Volume of NaOH = moles of NaOH / molarity of NaOH

Volume of NaOH = 0.0044 moles / 0.1052 M

Volume NaOH = 0.0418 L = 41.8 mL

We need 41.8 mL of NaOH

6 0

2 years ago

Two important ways that energy is transported in the world around us is through __________ and _________.

lana66690 [7]

The two ways that energy can be transferred are by doing work and by heat transfer.

8 0

3 years ago

Read 2 more answers

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

IgorC [24]

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

= 99.7°C - 20°C

= 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

= 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

7 0

3 years ago

How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t

ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

= 2.75 x 10²³ molecules XeF₆.

8 0

2 years ago

Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.

attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

Equivalentes KOH:

0.100L * (1eq / L) = 0.100eq = 0.100moles

Gramos KOH -Masa molar: 56.1056g/mol-:

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

KOH 90%:

5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>

8 0

2 years ago