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3 years ago

21. A piece of metal with a a mass of 15.2 g is heated from 17°C to 42°C. In the process it absorbs 1362 J of

Chemistry

1 answer:

Sholpan [36]3 years ago

4 0

Answer:

3.58J/g°C is the specific heat of the metal

Explanation:

The specific heat of a material is defined as the energy that 1g of the material absorbs and produce the increasing in temperature in 1°C. The equation is:

Q = S*ΔT*m

Where Q is energy = 1362J

S is specific heat of the material

ΔT is change in temperature = 42°C - 17°C = 25°C

And m is the mass of the material = 15.2g

Replacing:

S = Q / ΔT*m

S = 1362J / 25°C*15.2g

<h3>3.58J/g°C is the specific heat of the metal</h3>

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Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

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2 years ago

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2 years ago

Can some atoms exceed the limits of the octet rule in bonding? If so, give an example.

harkovskaia [24]

Answer:

Yes. Example: Sulfur hexafluoride (SF₆) molecule

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3 years ago

20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as

shepuryov [24]

The reaction forms 98.76 g AlCl_3.

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

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Step 2. Calculate the moles of each reactant

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Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the limiting reactant

Calculate the moles of AlCl_3 we can obtain from each reactant.

From Al: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

From Cl_2: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

Cl_2 is the limiting reactant because it gives the smaller amount of AlCl_3.

Step 4. Calculate the mass of AlCl_3.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

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3 years ago

What is FeBr3 compound name?

inessss [21]

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3 years ago

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