Answer:

Explanation:
It is given that,
The radius of the inflated balloon is 7 cm.
We need to find the volume of air inside the balloon in milliliters.
The balloon is in the shape of a sphere whose volume is given by :

Hence, the volume of the air inside the balloon is
.
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
The nucleus (plural, nuclei) houses the cell's genetic material, or DNA, and is also the site of synthesis for ribosomes, the cellular machines that assemble proteins. Inside the nucleus, chromatin (DNA wrapped around proteins, described further below) is stored in a gel-like substance called nucleoplasm.
I believe it becomes gasses the hotter it gets
Answer:

Explanation:
<u>Relation between grams of Carbon and moles</u> :

<u>Multiply both sides by 2</u> :
⇒ 1 x 2 moles = 12 x 2 g
⇒ 2 moles = 24 g