1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.

You might be interested in
Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g. *
ycow [4]

Answer:

90%

Explanation:

Percentage yield = ?

Theoretical yield = 50g

Actual yield = 45g

To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (45 / 50) × 100

Percentage yield = 0.9 × 100

Percentage yield = 90%

The percentage yield of the substance is 90%

6 0
3 years ago
Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5
e-lub [12.9K]

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to pK_{a} of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and CH_{3}COO^{-} is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

pK_{a}(CH_{3}COOH)=4.74

Buffer A:    pH=4.74+log(\frac{0.10}{0.10})=4.74

Buffer B:   pH=4.74+log(\frac{0.30}{0.30})=4.74

Buffer C:   pH=4.74+log(\frac{0.10}{0.50})=4.04

So, both buffer A and buffer B has same pH value which is also equal to pK_{a} . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from pK_{a} . Therefore, buffer C has the lowest buffer capacity.

     

6 0
3 years ago
Which type of volcano forms when eruptions contain mostly dush ashes and cinders?
Alisiya [41]
Stratovolcanoes, or composite volcanoes, mostly erupt dust, ash, and cinders.
6 0
3 years ago
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
Other questions:
  • What system does the heart go to?
    6·1 answer
  • When air bags inflate, nitrogen gas is formed from sodium azide, along with solid sodium. what reaction category is this?
    7·1 answer
  • Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
    9·1 answer
  • 1.) How can atomic number and mass number be used to find the numbers of protons, electrons, and neutrons?
    5·1 answer
  • A plant extract can be obtained with organic solvents such as acetone or ethyl alcohol, and is composed of plant pigments such a
    14·1 answer
  • Write a balanced chemical equation for the complete combustion of C3H7BO3, a gasoline additive. The products of combustion are C
    15·1 answer
  • What is a valence electron ?
    8·2 answers
  • 2. Calculate the density of a metal that occupies 17.75 cm and has a mass of 342.93 g. [D = m/V]
    9·1 answer
  • Thermal energy will flow from the person to the desk
    13·2 answers
  • PLEASEEEE HELP MEEEEE
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!