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kvv77 [185]
3 years ago
12

H2SO4 + 2 NaOH →

Chemistry
1 answer:
Alex73 [517]3 years ago
4 0

Answer:

110.25grams

Explanation:

The balanced chemical equation is as follows:

H2SO4 + 2NaOH → Na2SO4 + 2H20

Based on the balanced chemical equation of the reaction given above, 1 mole of sulphuric acid (H2SO4) is required to react with 2 moles of sodium hydroxide (NaOH).

Hence, if 1 mol of H2SO4 is needed to react with 2 moles of NaOH

Then, 2.25mol of NaOH will be required to react with;

= 2.25/2

= 1.125mol of H2SO4

Using the formula, mole = mass/molar mass, we can convert the molar value of H2SO4 to its mass value.

Molar mass of H2SO4 = 1(2) + 32 + 16(4)

= 2 + 32 + 64

= 34 + 64

= 98g/mol

Therefore, mole = mass/molar mass

1.125 = mass/98

mass = 98 × 1.125

mass = 110.25grams of H2SO4

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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
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Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
  • Specific heat of water = 1.035 \frac{cal}{gC}

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

#SPJ1

7 0
2 years ago
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