<span>The mass of one mole of sodium bicarbonate (aka NaHCO3) is equal to 1 * 22.99g/mol + 1 * 1.00g/mol + 1 * 12.01g/mol + 3 * 16.00g/mol = 83.91g/mol. From this, we can convert 4.2g of NaHCO3 to moles by dividing by 83.91g/mol, to get 0.050 moles of sodium bicarbonate.</span>
Answer: The average atomic mass of the element = 88.242amu
Explanation:
The abundance of the first isotope is =35.5%
Atomic mass of first isotope = 68.9257
The average atomic mass of the first isotope =86.95amu X 35.5% =86.95amu X 0.355 =30.8725 amu
The abundance of the second isotope =64.5%
Atomic mass of the second isotope =88.95amu
The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu
Now the average atomic mass =30.8725 +57.37275 = 88.242amu
OR using the formulae
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100
8,824/100
=88.24amu
<span>the answer is
C. The bar for very low concentration is twice the height of the bar for medium concentration.
proof
</span>
<span>
</span>
<span>Medium--------------------15
</span>Very Low-------------------30 = 2<span /> x 15 (<span>Medium)
</span>
Explanation:
<em>Potas</em><em>sium</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomi</em><em>c</em><em> </em><em>radii</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>greate</em><em>r</em><em> </em><em>t</em><em>han</em><em> </em><em>t</em><em>hat</em><em> </em><em>of</em><em> </em><em>lithiu</em><em>m</em><em> </em><em>th</em><em>at</em><em> is</em><em> </em><em>why</em><em> </em><em>pot</em><em>assium</em><em> </em><em>is</em><em> </em><em>more</em><em> </em><em>reactiv</em><em>e</em><em> </em><em>than</em><em> </em><em>lithi</em><em>um</em><em>.</em>
<em>tHx</em><em> </em><em>fOr</em><em> pOinTs</em><em>.</em><em>.</em><em>.</em>
