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creativ13 [48]
3 years ago
13

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

<em>Where Q is energy = 1362J</em>

<em>S is specific heat of the material</em>

<em>ΔT is change in temperature = 42°C - 17°C = 25°C</em>

<em>And m is the mass of the material = 15.2g</em>

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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<em>2. Plug in values. </em>V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)}  = 7.14 mL

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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]

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Yes. Example: <u>Sulfur hexafluoride (SF₆) molecule</u>

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According to the octet rule, elements tend to form chemical bonds in order to have <u>8 electrons in their valence shell</u> and gain the stable s²p⁶ electronic configuration.

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Whereas, there are <u>12 electrons around the central sulfur atom</u> in the SF₆ molecule. Therefore, <u>sulfur does not follow the octet rule.</u>

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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 <em>limiting reactant problem</em>.

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

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

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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 <em>limiting reactant</em>  

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

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

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<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

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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