Answer is: 0,453 <span> moles of oxygen will react </span><span>with 0.3020 moles of carbon(IV) oxide</span><span>.
</span>n(CO₂) = 0,302 mol.
From chemical reaction: n(CO₂) : n(O₂) = 4 : 6.
n(O₂) = 6 · 0,302 mol ÷ 4.
n(O₂) = 0,453 mol.
n - amount of substance.
Answer:
1.54 atm
Explanation:
By Dalton's Law Of partial pressure,
Total Pressure = Sum of all partial pressures
So,P= P1 + P2 + P3
Therefore, P=0.23+0.42+0.89
=1.54 atm
Answer:
a) 1 x 10^-11 mol/L
b) 1 x 10^-6 mol/L
c) 1 x 10^-5 fewer H+ ions
Explanation
pH stands for Power of Hydrogen, the more acidic a substance is, the more H+ ions it has rendering the substance acidic. a pH of 1 means the concentration of H+ ions is 1 x 10^-1. A pH of 7 means the concentration of H+ ions is 1 x 10^-7 and so on.
10^-11 has 10^-5 more H+ ions than 10^-6
Hope this helps :)
Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Answer:
25 grams
Explanation:
You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.
