Answer:
D. 15.8atm
Explanation:
Given parameters:
Initial pressure = 13atm
Initial temperature = 34°C = 34 + 273 = 307K
Final temperature = 100°C = 100 + 273 = 373K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.
The expression is shown mathematically below;
=
P and T pressure and temperature values
1 and 2 are initial and final states
Insert the parameters and solve for T₂;
=
P₂ = 15.8atm
Expanding in powers 4FCsixteen.
<h3>What is hexadecimal system?</h3>
Hexadecimal is an easy route for addressing twofold. It is vital to take note of that PCs don't utilize hexadecimal - it is utilized by people to abbreviate parallel to an all the more effectively reasonable structure. Hexadecimal Number System is generally utilized in Computer programming and Microprocessors. It is likewise useful to portray colors on pages. Every one of the three essential tones is addressed by two hexadecimal digits to make 255 potential qualities, hence bringing about in excess of 16 million potential tones. The primary benefit of utilizing Hexadecimal numbers is that it utilizes less memory to store more numbers, for instance it stores 256 numbers in two digits while decimal number stores 100 numbers in two digits. This number framework is likewise used to address Computer memory addresses.
Learn more about hexadecimal system, refer:
brainly.com/question/21751836
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Answer:
<u>Oxidation state of Mn = +4</u>
Explanation:
Atomic mass of Mn = 55g/mol
From Faraday's law of electrolysis,
Electrochemical equivalent = 
i.e Z =
=
= 0.0001424 g/C
But Equivalent weight, E = atomic mass ÷ valency = Z × 96,485
⇒
= 0.0001424 × 96,485
<u>∴ Valency of Mn = +4</u>
Answer:
alright that sounds like a recipe i gota try
Answer:The nucleus contains two types of subatomic particles,
protons and neutrons.
Explanation:The protons have a positive electrical charge and the neutrons have no electrical charge. A third type of subatomic particle, electrons, move around the nucleus.