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Setler79 [48]
3 years ago
6

What is the the heat of reaction, delta, in kJ/mol?

Chemistry
1 answer:
wariber [46]3 years ago
5 0

Answer:

b) -25 kJ/mol.

Explanation:

Hello there!

In this case, since the general definition of the enthalpy of reaction involves the subtraction of the energy of the products and the energy of the reactants:

\Delta H=H_{prod}-H_{react}

Thus, since the graph shows that the energy of the products is 20 kJ/mol and that of reactants 45 kJ/mol, we will obtain:

\Delta H=20kJ/mol-45kJ/mol\\\\\Delta H=-25kJ/mol

Which means it is exothermic.

Regards!

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Water beads up on waxy surfaces because of a degree of adhesion with the surface. True or False
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How many grams of Br2 are needed to completely convert 15.0 g Al to AlBr3 ?
wariber [46]

Answer:

133 g

Explanation:

Step 1: Write the balanced equation

2 Al(s) + 3 Br₂(l) ⇒ 2 AlBr₃(s)

Step 2: Calculate the moles corresponding to 15.0 g of Al

The molar mass of aluminum is 26.98 g/mol. The moles corresponding to 15.0 g of Al are:

15.0g \times \frac{1mol}{26.98g} = 0.556mol

Step 3: Calculate the moles of Br₂ that react with 0.556 moles of Al

The molar ratio of Al to Br₂ is 2:3. The moles of bromine that react with 0.556 moles of aluminum are:

0.556molAl \times \frac{3molBr_2}{2molAl} = 0.834molBr_2

Step 4: Calculate the mass corresponding to 0.834 moles of Br₂

The molar mass of bromine is 159.81 g/mol. The mass corresponding to 0.834 moles of Br₂ is:

0.834mol \times \frac{159.81g}{mol} = 133 g

8 0
3 years ago
Which properties are characteristic of group 2 elements at stp
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Why is the reaction mixture extracted with sodium bicarbonate?
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Vinegar which is an acid
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4 years ago
A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and
Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

0.03337 M=\frac{n}{0.02328 L}

n = 0.03337 M\times 0.02328 L=0.0007768 mol

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

\frac{1}{1}\times 0.0007768 mol=0.0007768 mol chloride ions.

In one mole of aldrin there are 6  moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

\frac{0.0007768 mol}{6}=0.0001295 mol

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

\frac{0.04726 g}{0.1064 g}\times 100=44.41\%

8 0
3 years ago
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