Answer:
D. 0.3 M
Explanation:
NH4SH (s) <--> NH3 (g) + H2S (g)
Initial concentration 0.085mol/0.25L 0 0
Change in concentration -0.2M +0.2 M +0.2M
Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M
concentration
Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M
K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M
<em>answer:</em><em> </em><em>option </em><em>d </em><em>(</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>H </em><em>+</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>O</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
Answer:
14.434 r.a.m.
Explanation:
- The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.
∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)
atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.
atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.
1) is chemical Bonds
3) Conservation of mass
5) compound
hope i helped on the ones i could answer
