Sorry I don’t get the question Buh I’m sure you’ll get it eventually
Answer:
C. the use of hydrogen gas as an electron donor.
Explanation:
Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.
Answer:
Average atomic mass = 51.9963 amu
Explanation:
Given data:
Abundance of Cr⁵⁰ with atomic mass= 4.34%
, 49.9460 amu
Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu
Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu
Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu
Average atomic mass = 51.9963 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100
Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100
Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100
Average atomic mass = 5199.632 / 100
Average atomic mass = 51.9963 amu
Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
Answer:
–0.13 Pa.m²
Explanation:
From the question given above, the following data were obtained:
Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²
Measurement (Pa.m²) =?
We can convert from Pa.mm² to Pa.m² by doing the following:
1 Pa.mm² = 1×10¯⁶ Pa.m²
Therefore,
–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²
–1.3×10⁵ Pa.mm² = –0.13 Pa.m²
Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².
