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pogonyaev
3 years ago
15

Use the molar heat capacity for aluminum from table 1 to calculate the amount of energy needed to raise the temperature of 260.5

g of aluminum from 0 c to 125 C
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
6 0
Unfortunately, you failed to include the table 1 from which the molar heat capacity of aluminum could have been obtained. However, as a general rule, the heat needed to raise the temperature of a certain substance by certain degrees is calculated through the equation,
                            H = mcpdT
where H is heat, m is mass, cp is specific heat capacity, and dT is change in temperature. From a reliable source, cp for aluminum is equal to 0.215 cal/g°C. Substituting this to the equation,
                               H = (260.5 g)(0.215 cal/g°C)(125°C - 0)
                                        H = 7000.94 cal
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2 years ago
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

7 0
3 years ago
Read 2 more answers
Consider the following reaction. Zn(s) 2agno3(aq) --> 2ag(s) zn(no3)2(aq)when 16. 2 g of silver was produced, _____ mole(s) o
Crazy boy [7]

The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Moles of silver = 16.2g / 107.8g/mol = 0.15mol

From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.

Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g

Hence required mass of silver nitrate is 25.48g.

To know more about mass & moles, visit the below link:

brainly.com/question/19784089

#SPJ4

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