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Strike441 [17]
3 years ago
12

Explain how stagnant sulfureted water may greatly accelerate metallic corrosion. (See the Email Link in Moodle) | (7 marks)

Chemistry
1 answer:
slavikrds [6]3 years ago
6 0

Answer:

sulfur promotes oxide-reduction reactions.

Explanation:

In stagnant water, some solutes tend to precipitate. When Sulfur precipitate and touch a metal, Sulfur is being reduced and the metal is oxidated. This depends of potential redox of each element.  

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You have been given two samples of water. One is from the atlantic ocean and the other is from lake superior. Describe how you w
lbvjy [14]

Answer:

i think by their salinity (the saltiness or amount of salt dissolved in a body of water)

Explanation:

Water salinity is measured by passing an electric current between the two electrodes of a salinity meter in a sample of water.

5 0
3 years ago
What does the mass number of an atom represent?.
mamaluj [8]

Answer:

the sum of the numbers of protons and neutrons present in the nucleus of an atom.

8 0
2 years ago
"calculate the ratio of the velocity of hydrogen molecules to the velocity of carbon dioxide molecules at the same temperature"
Ne4ueva [31]

Answer: 1:4.69

Explanation:

The ratio can be expressed as:

Ua/Ub= √(Mb/Ma)

Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2

Mb is the molecular mass of CO2 = 44

Therefore

Ua/Ub= √(44/2)

Ua/Ub = 4.69

Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69

which implies hydogen is about 4.69 times faster than carbon dioxide.

8 0
4 years ago
Why glucose and fructose produce same shaped products with phenylhydrazine?
NemiM [27]

During the reaction of glucose and fructose with excess phenylhydrazine to form osazone, only the C-1andC-2 atoms of glucose and fructose participate in the reaction. The rest of the molecule remains intact. Hence, glucose and fructose produce the same osazone.

6 0
3 years ago
The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
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