Answer:
Publishing results of research projects in peer-reviewed journals enables the scientific and medical community to evaluate the findings themselves. It also provides instructions so that other researchers can repeat the experiment or build on it to verify and confirm the results.
Answer:
The P-H bonds are more polar than the N-H bonds.
Explanation:
Phosphine is a polar molecule with non-polar bonds. The phosphorus atom is bonded to three hydrogen atoms and the phosphorus atom has a lone pair of electrons. Since hydrogen and phosphorus are equal in electronegativity, it implies that they attract the shared pairs of electrons the same amount,hence bonding electrons are shared equally making the covalent bonds non-polar.
The lone pair of electrons on phosphorus causes the molecule to be asymmetrical with respect to charge distribution this is why the molecule is polar even though the are non-polar bonds in the molecule.
Looking at the values of electro negativity stated in the question, one can easily see that the difference in electro negativity between nitrogen and hydrogen is 0.9 while the difference in electro negativity between phosphorus and hydrogen is zero. It is clear that NH3 is naturally more polar than PH3 since each individual N-H bond in NH3 is a polar bond while the individual P-H bonds in PH3 are nonpolar.
Answer:
vertebrate and invertebrates
Explanation:
All Vertebrates have a skeletal structure with a spinal backbone and well developed internal skeleton made up of bones and cartilage while Invertebrates have no backbone.
Vertebrate include fish, amphibians, reptiles, birds and mammals. Invertebrates include worms and jellyfish, spiders and crabs.
<u>common characteristics among vertebrates and invertebrates </u>
○ multi-cellular.
○ heterotrophs (cannot make their own food) and must get their energy by consuming plants or other animals.
○ obtain food and oxygen for energy,
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
