ωєℓℓ тнє ρнσѕρнσяι¢ α¢ι∂ мσℓє¢υℓєѕ αттα¢н тσ тнє мσℓє¢υℓєѕ σf тнє мιℓк, αи∂ тнαт ιи¢яєαѕєѕ тнє ∂єиѕιту αи∂ тнєи ѕєρєяαтєѕ тнєм fяσм тнє яєѕт σf тнє ℓιqυι∂ ιи ιт. тнє яємαιи∂єя σf тнє ℓιqυι∂ѕ,иσω нανιиg ℓєѕѕ ∂єиѕιту тнαи тнє ρнσѕρнσяι¢ α¢ι∂ѕ & тнє мιℓк мσℓє¢υℓєѕ, ѕσ ιт ιт иσω fℓσαтѕ σи тσρ.
нσρє ι ¢συℓ∂ нєℓρ уσυ.
I believe it lies in the 3rd Quadrent
Answer: pOH = 3.13
Ba(OH)2 is a very basic substance. The general formula for pOH is - log(OH)
Barium Hydroxide produces 2 moles of OH for every mole of Ba(OH)2 present in the solution.
0.00037 M = 3.7 * 10^-4 Ba(OH)2 will produce 2 *0.00037 M of OH-
OH- = 2* 0.00037 = 0.00074
pOH = - log(0.00074) = 3.13
I think the answer to this question will be +7kj/mol
Answer:
<span>Given mass of C is 32.0 grams
</span><span>Given mass of H is 8.0 grams
</span>Molar mass of C is 1212 g/mol
Molar mass of H is 1.01 g/mol
Thus
<span><span><span><span>32.0 grams of C </span><span>8.0 grams of H </span></span><span><span>→<span>3212</span>=2.66 moles of C</span><span>→<span>81.01</span>=7.92 moles of H</span></span></span><span><span><span>32.0 grams of C </span><span>→<span>3212</span>=2.66 moles of C</span></span><span><span>8.0 grams of H </span><span>→<span>81.01</span>=7.92 moles of H</span></span></span></span>
<span>When we divide 7.92 by 2.66 we obtain 2.977 which is approximately 3. This means that the ratio of atoms of </span><span>CC</span><span> to the atoms of </span><span>HH</span><span> is 1:3.</span>
<span>Thus, empirical formula for the compound is </span><span><span>C<span>H3</span></span><span>C<span>H3</span></span></span>.
<span>Molar mass of </span><span><span>C<span>H3</span></span><span>C<span>H3</span></span></span><span> is </span><span><span>1⋅12+3⋅1.01=15.03</span><span>1⋅12+3⋅1.01=15.03</span></span><span>. Since molar mass of the compound we have to find is </span><span>3030</span><span> g/mol we have tu multiply subscripts by 2.</span>
<span>Thus, final compound is </span><span><span><span>C2</span><span>H6</span></span><span><span>C2</span><span>H6</span></span></span>.
