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alexdok [17]
3 years ago
7

1. Though Florida receives a lot of rainfall, only 1 percent remains available for people to use or drink. What happens to the r

est of the water? 2. Because Florida has such a large underground water system, which four (4) other states receive drinking water from Florida? 3. Name one reason people in Florida wanted to get rid of the water in certain areas. 4. What was one negative result of the changes people made to get rid of the water? 5. What is one thing that has been done to try and improve those mistakes?
Chemistry
1 answer:
timama [110]3 years ago
8 0
<span>I am sure these answers are correct cause I've checked all of them. Here they are : 

1) The thing which might h</span>appen to the rest of the water is that 61% of that water leaves the ground through evapotranspiration and <span>38% will flow into the Atlantic Ocean. 
</span>
2) The four states which  receive drinking water from Florida are : <span> South Carolina, Georgia, Alabama, Mississippi 

3) The</span> reason why  people in Florida wanted to get rid of the water in certain areas is abundance of wetlands. I think there is one more, if it is acceptable I suggest you using this - frequent flooding.<span>

4) </span><span> The negative result of the changes which people made to get rid of the water is the first sign of </span>drought.

5) The thing that has been done to try and improve those mistakes was creating a tool which looked like a dam to prevent dfficulties with <span>navigating the state.

Hope that helps.</span>
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2. How many grams of NaCl are required to prepare 0.40 L of a 0.75 M solution?
Yuri [45]

The mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g. Details about mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

However, the number of moles of a solution must be initially calculated by using the following formula:

molarity = no of moles ÷ volume

no of moles = 0.75 × 0.40

no of moles = 0.3 moles

mass of NaCl = 0.3 × 58.5 = 17.55g

Therefore, the mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g.

Learn more about mass at: brainly.com/question/19694949

#SPJ1

4 0
2 years ago
What was the robotic vehicle Opportunity designed to do?
mestny [16]
I believe the answer is C
6 0
3 years ago
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Electrons are attracted to ________________. protons other electrons neutrons all the particles
rjkz [21]
Protons .because electron has negative charge and proton has positive charge.
6 0
3 years ago
Identify and name the global<br> wind belts.<br> A<br> B<br> C<br> D<br> E<br> F<br> G
Finger [1]

Answer:

A. Polar Easterlies

B. Westerlies

C. Northeast Trades

D. Southeast Trades

E. Westerlies

F. Polar Easterlies

G. Hadley cell

Explanation:

I am pretty sure this is right. Hope that helps

8 0
3 years ago
The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment
astraxan [27]

Answer:

The activation energy is =8.1\,kcal\,mol^{-1}

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_{A}=kC_{A}

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^{3}

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_{o}=5dm^{3}s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]

Rearrange the formula is as follows.

k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)

The feed has Pure A, mole fraction of A in feed y_{A_{o}} is 1.

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

Substitute the all given values in equation (1)

k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}

Therefore, the rate constant in case of the plug flow reacor at 300K is1.2s^{-1}

The rate constant in case of the CSTR can be calculated by using the formula.

\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed y_{A_{o}} is 0.5

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacted.

=0.5(2-1)=0.5

Substitute the all values in formula (2)

\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}

Therefore, the rate constant in case of CSTR comes out to be 2.8s^{-1}

The activation energy of the reaction can be calculated by using formula

k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}

Substitute the all values.

=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}

=8.1\,kcal\,mol^{-1}

Therefore, the activation energy is =8.1\,kcal\,mol^{-1}

8 0
3 years ago
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