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3 years ago

Do organs of different species resemble each other?

1 answer:

6 0

<span>This really depends on how closely related the species are. Species from vastly unrelated taxonomic groups are likely to have organs that differ substantially. Think for example of the compound eye of a spider and the eye of a human, or the bones of a fish compared to the cartilage of a shark. These are examples of species that are not closely related at all. Then think of a chimpanzee and a human. The organs of both species are very similar in form and function as they are closely related. </span>

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Can Iron replace Hydrogen in the following reaction?: Fe(s) + HCl(aq) ---> H2(g) + FeCl2(aq)

Elenna [48]

It can because Iron is more reactive than Hydrogen and a more reactive metal always replaces a less reactive one in a single replacement reaction

3 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago

How does matter change from one state to another?

Rzqust [24]

Heating up, mostly. Solid can be heated to liquid, then to gas, then eventually to plasma.

4 0

3 years ago

How many grams of lead(II) iodide are produced from 6.000 moles of NaI according to the balanced equation: Pb(NO3)2 + 2 NaI à 2

marta [7]

Answer:

mass PbI₂ formed = 1383 grams

Explanation:

Pb(NO₃)₂ + 2NaI => 2NaNO₃ + PbI₂(s)

6 mol NaI => 1/2(6 mol) PbI₂ = 3 mol PbI₂ x 461.01 g/mol = 1383.03 grams ≅ 1383 grams (4 sig. figs.)

7 0

2 years ago

U-235 undergoes many different fission reactions. For one such reaction, when U-235 is struck with a neutron, Ce-144 and Sr-90 a

Lostsunrise [7]

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

The given reaction is $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{144}_{58}Ce + ^{90}_{38}Sr + x^{1}_{0}n + y^{0}_{-1}\beta$

Now, we balance the mass on both reactant and product side as follows.

235 + 1 = $144 + 90 + (x \times 1) + (y \times 0)$

236 = 234 + x

x = 236 -234

= 2

So, now we balance the charge on both reactant and product side as follows.

92 + 0 = $58 + 38 + (x \times 0) + (y \times -1)$

92 = 96 - y

y = 4

Thus, we can conclude that there are 2 neutrons and 4 beta-particles are produced in the given reaction.

Therefore, reaction equation will be as follows.

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{144}_{58}Ce + ^{90}_{38}Sr + 2^{1}_{0}n + 4^{0}_{-1}\beta$

8 0

3 years ago