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aev [14]
3 years ago
5

Anabolic reactions _______ bonds, whereas catabolic reactions __________ bonds. A. decrease; increase. B. break; make C. weaken;

strengthen D. loosen; tighten. E. make; break
Chemistry
1 answer:
nexus9112 [7]3 years ago
6 0

Answer:

The corrext answer is E. make; break

Explanation:

In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures

The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds

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The answer is +1! Have a great day!
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What do cations and anions form?
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How many molecules are contained in a 3.36 mole sample of NH3?
denpristay [2]

3.36 \times (6.02 \times  {10}^{23} ) = 2.0227 \times  {10}^{24}
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An organism's _______ describes its genetic composition. An organism's _______ describes its appearance or observable characteri
Kisachek [45]
B. 
<span>genotype; phenotype

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7 0
3 years ago
Read 2 more answers
The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) th
seropon [69]

Answer:

(a) a = 5.08x10⁻⁸ cm

(b) r = 179.6 pm  

Explanation:

(a) The lattice parameter "a" can be calculated using the following equation:

\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}      

<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>N_{A}<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>

Hence the lattice parameter is:  

a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}

a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm

(b) We know that the lattice parameter of a FCC structure is:

a = \frac{4r}{\sqrt{2}}

<em>where r: is the atomic radius of Th</em>

Hence, the atomic radius of Th is:

r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm    

I hope it helps you!    

4 0
3 years ago
Read 2 more answers
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