Attached to this answer is the format of Isotope Notation that you can use for future reference. <em>(Please open)</em>
There are
8 Protons. The Atomic Number is the same number of an element's proton.
If you can see in the format, the mass number is calculated by adding the atomic number/protons and neutrons.
Mass number = 8 + 11Mass number = 19The image of the final answer is attached as well.
Answer:
4,313.43 mmHg is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg
Explanation:
Boyle's law says:
"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." This means that if the quantity of gas and the temperature remain constant, the product of the pressure for the volume always has the same value.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
If you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment and you vary the volume of gas to a new value V2, then the pressure will change to P2, it will be true:
P1 * V1 = P2 * V2
In this case:
- V1=0.335 L
- P1= ?
- V2= 1700 mL= 1.7 L (Being 1 L=1000 mL)
- P2= 850 mmHg
Replacing:
P1*0.335 L=850 mmHg*1.7 L
Solving:

<u><em>P1=4,313.43 mmHg</em></u>
<u><em>4,313.43 mmHg is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg</em></u>
Answer:
<h2>My My name: CORN CORNELIUS CORNWALL</h2><h2>My Age: 209374329 years old</h2><h2>Fav song : Baby by justin bieber</h2><h2>most legendary thing that i got : club penguin membership</h2>
<u>M</u><u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>e</u><u> </u>is a carbon compound which undergoes combustion to <em><u>release energy</u></em> and form bi production which are <u>Carbon</u><u> </u><u>dioxide</u><u> </u>( CO2 )<u> </u><u>and</u><u> </u> <u>W</u><u>ater</u> ( H20 ).
the balanced chemical equation for the reaction is : -
Answer:
carbon
Explanation:
Atomic radius represents the distance from the nucleus to the outer shell of an element.