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telo118 [61]
3 years ago
5

Sodium carbonate is a reagent that may be used to standardize acids. In such a standardization it was found that a 0.498-g sampl

e of sodium carbonate required 21.3 mL of a sulfuric acid solution to reach the endpoint for the reaction.
Na2CO3 (aq) + H2SO4 (aq) → H2O (l) + CO2 (g) + Na2SO4 (aq)


What is the molarity of the H2SO4?
Chemistry
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

The molarity of H2SO4 is 0.221 M

Explanation:

Step 1: Data given

Mass of sodium carbonate (Na2CO3) = 0.498 grams

Molar mass Na2CO3 = 105.99 g/mol

Volume of sulfuric acid (H2SO4) = 21.3 mL = 0.0213 L

Step 2: The balanced equation

Na2CO3 (aq) + H2SO4 (aq) → H2O (l) + CO2 (g) + Na2SO4 (aq)

Step 3: Calculate moles Na2CO3

Moles Na2CO3 = 0.498 grams / 105.99 g/mol

Moles Na2CO3 = 0.00470 moles

Step 4: Calculate moles H2SO4

For 1 mol Na2CO3 we need 1 mol H2SO4 to react and produce 1 mol H2O, 1 mol CO2 and 1 mol Na2SO4

For 0.00470 moles Na2CO3 we need 0.00470 moles H2SO4 to react

Step 5: Calculate molarity of H2SO4

Molarity H2SO4 =moles H2SO4 / volume

Molarity H2SO4 = 0.00470 / 0.0213 L

Molarity H2SO4 = 0.221 M

The molarity of H2SO4 is 0.221 M

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A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
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Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0
3 years ago
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