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4 years ago

How do we turn petroleum into energy we use?

1 answer:

7 0

You don't "turn" it into energy; petroleum HAS stored energy (chemical energy).However, you can turn it into ANOTHER TYPE OF ENERGY; usually this is done by burning the petroleum, and using it to drive machinery.

Since burning fuels is wasteful (the efficiency is limited, in theory, to the Carnot efficiency of a heat engine), other options are being explored, such as chemical reactions in a fuel cell. But such technology is not yet used on a large scale.

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Identify each of the following:

zaharov [31]

Most metallic: Ge

Largest atomic radius: Bi

Highest ionization energy: Cl

Could you mark me as brainliest?

7 0

3 years ago

Which Domains would you find single cell organisms

kogti [31]

Archaea and Bacteria

8 0

3 years ago

Marat540 [252]

Answer:

The correct option is A) The light moths will be captured by predators more easily than the dark moths, and the population of dark moths will rise.

Explanation:

As we can see, the colour of the trees do not match with the light moths. Instead, the colour of the trees resembles the dark moths. This adaptation will work best for the dark moths to protect it from its predators. The light moths lack this adaptation and can easily be attacked by the predators. Hence, the population of the dark moth will increase as they are better adapted to live in such an environment.

3 0

3 years ago

Balance the following equation in acidic conditions Phases are optional. S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

jonny [76]

To balance the the chemical reaction, the number of moles per element is balance is both side of the reaction and also the charge in both sides of the reation. to balnce the reaction:

S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

2S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+ + e

3 0

3 years ago

For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

3 0

3 years ago