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AlladinOne [14]
4 years ago
6

A.) During an experiment, a student adds 2.90 g of CaO to 400.0 mL of 1.500 M HCl. The student observes a temperature increase o

f 6.00 °C. Assuming the solution\'s final volume is 400.0 mL, the density if 1.00 g/mL, and the heat capacity is 4.184 J/(g·°C), calculate the heat of the reaction, ΔHrxn.
B.) During an experiment, a student adds 1.05 g of calcium metal to 200.0 mL of 0.75 M HCl. The student observes a temperature increase of 17.0 °C for the solution. Assuming the solution\'s final volume is 200.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J/(g·°C), calculate the heat of the reaction, ΔHrxn.
C.) Write the balanced chemical equation for the standard enthalpy of formation of magnesium oxide. Include the phase labels. The arrow has been provided for you. Do not include charges.
Chemistry
1 answer:
Nostrana [21]4 years ago
5 0

Answer:

a.  1.00 x 10⁴ J = 10.0 kJ

b.  1.42 x 10⁴ J = 14.2 kJ

Explanation:

Given the change in temperature during the reaction and assuming the volume of water and density  remains constant, the change in enthalpy for the reaction will be given by

ΔHxn = Q = mCΔT  where,

                               m= mass of water

                               C= specific heat of water, and

                               ΔT= change in temperature

a. mH₂O = 400.0 mL x 1.00g/mL = 400.00 g

Q = ΔHrxn = 400.00g x 4.184 J/gºC x 6.00 ºC = 1.00 x 10⁴ J = 10.0 kJ

b. mH₂O = 200.0 mL x 1.00 g/mL = 200.0 g

Q = ΔHrxn = 200.00 g x 4.184 J/gºC x 17.0ºC =  1.42 x 10⁴ J = 14.2 kJ

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A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and
zlopas [31]

Answer:

37.1°C.

Explanation:

  • Firstly, we need to calculate the amount of heat (Q) released through this reaction:

<em>∵ ΔHsoln = Q/n</em>

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

  • We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>

6 0
3 years ago
What does a metal's position in an activity series indicate about the chemical properties of that particular metal?
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The closer to the top the metal is in the list, the more active the metal is and the stronger a reducing agent the metal is. When two different metals are involved in a redox reaction, the metal higher in the list will be oxidized and give up electrons that will reduce the cation of the less active metal.
3 0
3 years ago
a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v
Rashid [163]

<u>Concentration of NaOH = 0.148 molar, M</u>

<u>Concentration of H3PO4 = 0.172 molar, M</u>

<u></u>

Concentration x Volume  will give the number of moles of solute in that volume.  C*V = moles

Concentration  has a unit of (moles/liter).  When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution:  (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl.  The reaction is:

  NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each.  We just determined that the  moles of HCl used was 0.00370 moles.  Therefore, the 25.0 ml solution of NaOH had the same number of moles:  0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters).  The concentration of NaOH is therefore:  

    <u>(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M</u>

====

The phosphoric acid problem is handled the same way, but with an added twist.  Phosphoric acid is H3PO4.  We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4.  But now the acid has three hydrogens that will react.  The balanced equation for this reaction is:

  H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need <u><em>three times</em></u> the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

  (0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml.  The concentration is therefore:

     (0.00172 moles H3PO4)/(0.010 liters H3PO4)

<u>Concentration of H3PO4 = 0.172 molar, M</u>

 

5 0
1 year ago
Which produces the most energy? A. passing 5 kg of water through a hydroelectric power plant. B. fissioning 5 kg of uranium. C.
blagie [28]

Answer:

D. fusing 5 kg of hydrogen.

Explanation:

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3 years ago
A quantity must be divided by multiples of ten when converting from a larger unit to a smaller unit.
seraphim [82]

Answer:It is false

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I took a quiz with this question in it and I chose true but I got it wrong

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