Answer:
i need a picture to solve
Explanation:
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:
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The mole fraction of solute in a 3.87 m aqueous solution is 0.0697
<h3>
calculation</h3>
molality = moles of the solute/Kg of the solvent
3.87 m dissolve in 1 Kg of water= 1000g
find the moles of water= mass/molar mass
that is 1000 g/ 18 g/mol= 55.56 moles
mole of solute = 3.87 moles
mole fraction is = moles of solute/moles of solvent
that is 3.87/ 55.56 = 0.0697
Answer:
this showed that the cathode rays traveled in straight line
An image is attached depicting the Lewis structure of HClO with the atoms arranged in that order. The result is a single bond between each atom. We can calculate the formal charge of each atom with the following formula:
Formal charge = (Valence electrons) - (electrons shared in bonds) - (non-bonded electrons)
H: Formal charge = 1 - 1 - 0 = 0
Cl: Formal charge = 7 - 2 - 4 = +1
O: Formal charge = 6 - 1 - 6 = -1
This structure is not the actual way these atoms arrange. The atoms will actually arrange in the order of H-O-Cl to form hypochlorous acid with a single bond between the three atoms and formal charges of 0 on each atom.