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aalyn [17]
3 years ago
5

Y’all answer as many as u can they just need to be correct thx!!!!

Chemistry
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

10. Embryology is the branch of biology and medicine concerned with the study of embryos and their development.

11. I think the answer is, Both humans and whale embryos have a tail and a backbone (I'm not sure).

12. Basilosaurid, any member of the family Basilosauridae, an early group of whales that lived from the middle Eocene to the late Oligocene Epoch (about 41 million to 23 million years ago).

13.  Protocetids had long snouts, large eyes, and their nasal opening was farther caudally than in earlier archaeocetes.

14. Two ways we know that Maiacetus was a whale re:

1) Their skeletons have all been found among fossils of sea-creatures which suggests they lived in the sea.

2) Their hip bones were strong enough to walk on land, but their short legs combined with long flat fingers and toes suggest they would have had difficulty walking.

15. The name Maiacetus means Mother Whale. Maiacetus is significant because of how well the skeletons were preserved.

Hope this helps! :D

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Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air
Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s

b) for completion of 32% of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-32}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}

t=44.3s

c) for completion of 81 % of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-81}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}

t=191s

4 0
3 years ago
Boyles Law P1V1 = P2V2
arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

P_1V_1=P_2V_2

Let the initial pressure of the balloon be P, i.e. P_1=P, initial volume be V, i.e. V_1=V. The pressure is then halved, i.e. P_2=\frac{P}{2}

P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V

Therefore the balloon volume will increase to twice their original volume.

3 0
3 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
A baseball has a mass of 135 grams and a softball has a mass of 270 grams. In which of the following situations would they have
liraira [26]

Answer:

The baseball is thrown twice as fast as the softball in the same direction.

Explanation:

8 0
3 years ago
Choose the FALSE statement.
Mariana [72]

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.
5 0
3 years ago
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