The chemical equation is:
<span>HCO3^ -> H^+ + CO3^2- </span>
We know that the formula
for Ka is:
Ka = [H^+][CO3^2-]/[HCO3^-]
log Ka = log[H^+] + log[CO3^2-]/[HCO3^-]
pKa = pH - log[CO3^2-]/[HCO3^-]
log[CO3^2-]/[HCO3^-] = pH - pKa = 10.75 - 10.329 = 0.421
<span>[CO3^2-]/[HCO3^-] = Antilog (0.421) = 2.636 </span>
Answer:
<span>2.636</span>
Potential because it hasnt released yet
Carbon atoms oxidation number is +2.Hydrogen's oxidation number is +1.Oxygen's oxidation number is -2.
<h3>Answer : 2C + 3H2 -> C2H6</h3>
<h3>Step and explanation</h3>
Condition : graphite + hydrogen gas
Chemical formula : C + H2
Reaction formula :
C + H2 -> C2H6
#First, you see the number of atom
•Product
C = 1
H = 2
•Reactant
C = 2
H = 6
<em>#</em><em>N</em><em>o</em><em>t</em><em>e</em><em> </em><em>that</em><em>,</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>chemical</em><em> </em><em>reaction</em><em> </em><em>both</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>each</em><em> </em><em>atom</em><em> </em><em>in</em><em> </em><em>product</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>equal</em><em> </em><em>in</em><em> </em><em>reactant</em><em> </em><em>so</em><em>,</em><em> </em><em>you</em><em> </em><em>think</em><em> </em><em>what</em><em> </em><em>number</em><em> </em><em>you</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>put</em><em> </em><em>Infront</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>chemical</em><em> </em><em>formula</em><em> </em><em>so</em><em> </em><em>that</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>atom</em><em> </em><em>equal</em>
2C + 3H2 -> C2H6
#Now check again
•Product
C = 2
H = 6
•Reactant
C = 2
H6
#Now, its equal. The bold one is the balance chemical reaction
<h3>Note (just incase u dont know)</h3>
Reactant -> product (the thing you get after reaction)
<span>Answer: 17.8 cm
</span>
<span>Explanation:
</span>
<span>1) Since temperature is constant, you use Boyle's law:
</span>
<span>PV = constant => P₁V₁ = P₂V₂
</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:
</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>
<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:
</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³
</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>