Condensed electron configuration of :
Ga = [Ar] 3d¹⁰ 4s² 4p¹.
O = 1s² 2s² 2p⁴
Gallium belongs to group 13 of the periodic table with 3 valence electrons. Oxygen (Atomic number = 8 and electronic configuration = 2,6) belongs to group 16 of the periodic table with 6 valence electrons. To achieve the octet stability criteria 2 gallium atoms will donate their 3 electrons each to 3 oxygen atoms.
The 3 oxygen atoms due to their high electron attracting ability will accept those electrons resulting in Ga3+ cation and O₂⁻ anion.
Thus, the formula of their compound if Ga₂O₃.
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Low life?- i’m pretty sure
<span>HCl<span>(aq)</span>+NaOH<span>(aq)</span>→NaCl<span>(aq)</span>+<span>H2</span>O<span>(l)</span></span>
As you can see here, one mole of acid neutralizes one mole of base.
We use the concentration equation, which states that,
<span>c=<span>nv</span></span>
<span>
<span>
<span>
n is the number of moles
</span>
<span>
v is the volume of solution
</span>
</span>
</span>
Rearranging for moles, we get,
<span>n=c⋅v</span>
So, we have:
<span><span>n<span>NaOH</span></span>=0.1 M⋅0.05 L</span>
<span>=0.005 mol</span>
Since one mole of acid neutralizes one mole of base, then we must have: <span><span>n<span>HCl</span></span>=<span>n<span>NaOH</span></span></span>.
And so,
<span><span>c<span>HCl</span></span>=<span><span>n<span>HCl</span></span><span>v<span>HCl</span></span></span></span>
<span>=<span><span>0.005 mol</span><span>0.03 L</span></span></span>
<span>≈0.17 <span>M</span></span>
Answer:
11 molecules of CH4.
23 atoms of C is the leftover.
Explanation:
Hello!
In this case, for the formation of methane:
We can see there is an excess of carbon based on their stoichiometry, because the needed amount of hydrogen gas molecules would be:
Thus, the formed molecules of methane are computed below:
In such a way, the leftover of carbon atoms are:
Best regards!
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
