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Leno4ka [110]
3 years ago
10

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

1.0°C, what is the specific heat of the substance?
Chemistry
1 answer:
Alexxandr [17]3 years ago
5 0
Specific heat capacity is the required amount of heat per unit of mass in order to raise teh temperature by one degree Celsius. It can be calculated from this equation: H = mCΔT where the H is heat required, m is mass of the substance, ΔT is the change in temperature, and C is the specific heat capacity.

H = m<span>CΔT
2501.0 = 0.158 (C) (61.0 - 32.0)

C = 545.8 J/kg</span>·°C
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If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
2 years ago
PLEASE HELP WITH EASY 8th GRADE SCIENCE!!
atroni [7]
I think c is the right answer
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3 years ago
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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

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3 years ago
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