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Leto [7]
3 years ago
8

The chart below lists data on four different projects designed to restore a wetlands habitat destroyed by human activity and a r

ecent hurricane.
The project that will be chosen will have to be low in cost, benefit the greatest number of species, and be popular with the local citizens whose taxes will sponsor it. Based on the data and the criteria given,which project will be chosen?

Chemistry
1 answer:
snow_lady [41]3 years ago
5 0

Answer:

Project 3.

Explanation:

Project 3's anticipated cost is 12 to 17 million dollars. It is a <em>lower </em>anticipated cost than Project 2 and Project 4, but <em>higher</em> than Project 1 by one million dollars at maximum cost anticipation. Additionally, the percentage of wildlife to benefit is 70-80%, which is <em>second</em> to the most wildlife to benefit which is Project 4 at 75-80%.

And finally, for community support for Project 3 - the chart lists it as high. This outclasses Project 2 and Project 4, but balances with Project 1. However, Project 1 costs 13 to 16 million dollars and <em>only</em> benefits 15-25% of wildlife.

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The stars are H and the circles are O by the way
shepuryov [24]

Answer:2 is correct

Explanation:

6 0
3 years ago
Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What
Leto [7]

Answer: The nuclide symbol of X is ^{231}_{90}\textrm{Th}

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}

For the given equation :

^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}

As the atomic number and mass number must be equal on both sides of the nuclear equation:

^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}

Thus the nuclide symbol of X is ^{231}_{90}\textrm{Th}

5 0
3 years ago
Which of the following is an oxidation-reduction reaction? a. HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) b. Pb(C2H3O2)2(aq) + 2 NaCl
saw5 [17]

Answer: Option (d) is the correct answer.

Explanation:

A reaction in which there occurs change in oxidation state of reacting species is known as an oxidation-reduction reaction.

(a)    HCl(aq) + LiOH(aq) \rightarrow LiCl(aq) + H_{2}O(l)

Will be written as:

H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow Li^{+} + Cl^{-} + H^{+} + OH^{-}

In this reaction, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(b)   Pb(C_{2}H_{3}O_{2})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s) + 2 NaC_{2}H_{3}O_{2}(aq)

Will be written as:

  Pb^{2+} + 2C_{2}H_{3}OO^{-} + 2Na^{+} + 2Cl^{-} \rightarrow Pb^{2+} + 2Cl^{-} + 2Na^{+} + 2C2H3OO^{-}

Similarly here,  there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(c)   NaI(aq) + AgNO_{3}(aq) \rightarrow AgI(s) + NaNO_{3}(aq)

Will be written as:

Na^{+} + I^{-} + Ag^{2+} + NO^{2-}_{3} \rightarrow AgI(s) + Na^{+} + NO^{2-}_{3}(aq)

Here, also there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(d)    Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)

So, here there occurs change in oxidation state of Mg from 0 to +2 and oxidation state of H changes from +1 to 0. Hence, it is an oxidation-reduction reaction.

Thus, we can conclude that Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g) is an oxidation-reduction reaction.

7 0
3 years ago
Which cell organelle is where proteins are made? A. nucleus B. ribosome C. vacuole D. mitochondrion
N76 [4]

Answer:

B. ribosome

Explanation:

In the ribosomes, the codons get paired with anti codons to create a polypeptide or protein.

5 0
4 years ago
Read 2 more answers
A heliox tank contains 32% helium and 68% oxygen. The total pressure in the tank is 475 kPa. What is the partial pressure of hel
Paladinen [302]

Answer:

152 kPa = Partial pressure O₂

Explanation:

Data by percent is the molar fraction . 100.

Molar fraction of Helium = 32/ 100 → 0.32

Molar fraction of O₂ = 68/100 → 0.68

Sum of molar fractions in a mixture = 1

0.68 + 0.32 = 1

If we apply the molar fraction, we can determine the partial pressure.

Mole fraction = Partial pressure / Total pressure

0.32 = Partial pressure O₂ / 475kPa →  0.32 . 475 kPa = Partial pressure O₂

152 kPa = Partial pressure O₂

8 0
3 years ago
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