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svetoff [14.1K]
3 years ago
9

An adult mosquitoes lifespan is about 16 days how many minutes is that

Chemistry
1 answer:
Delicious77 [7]3 years ago
5 0
1 day = 1, 440 minutes
16 days = (1, 440 × 16) days = 23, 040 minutes
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Suppose you mix one mole of sulfuric acid (H2SO4) with 1 mole of sodium hydroxide(NaOH).
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Answer & Explanation:

  • The neutralization of H₂SO₄ with NaOH is occurred according to the balanced equation:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O,</em>

It is clear that every 1.0 mol of H₂SO₄ needs 2 mol of NaOH to be neutralized completely.

<em>So, when you mix one mole of sulfuric acid with 1 mole of sodium hydroxide, there will be an excess of sulfuric acid.</em>

<em>Thus, the pH of the solution remain below 7.</em>

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3 years ago
Choose the correct option for the redox reaction below : 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) 4H2O(l) H+ (oxid
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Explanation:

8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

To determine if a substance was reduced or oxidized, the oxidation number at the reactant and product side is compared.

When oxidation occurs, an element loses electrons and its oxidation number increases (becomes more positive). When reduction occurs, an element gains electrons and its oxidation number decreases or is reduced (becomes more negative).

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MnO4- (oxidizing or reducing agent) Answer 2 Choose...

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Fe2+ (oxidizing or reducing agent) Answer 3 Choose...

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Fe2+ (oxidized or reduced) Answer 4 Choose...

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O in MnO4- (oxidized or reduced) Answer 5 Choose...

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Mn in MnO4- (oxidized or reduced) Answer 6 Choose...

There is a decreaase in oxidation number from +7 to +2. This means reduction.

6 0
4 years ago
A student finds that an unknown element reacts with elements from Group 2 (IIA). To which group does the unknown element most li
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A because a

8 0
3 years ago
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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

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