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JulijaS [17]
3 years ago
15

Name 4 invasive species and describe their effect on ecosystem.

Chemistry
1 answer:
Grace [21]3 years ago
6 0

Spotted Lantern: Destroying our trees

Fishers: Brought to lower numbers of certain species, but have dramatically lowered other species.

Zebra Mussels: Clog pipes

Emerald Ash Borer: Ruins ash trees that provide us with durable wood.

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Answer:

it was pillars

Explanation:

i guessed and it was right LOL

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6 0
2 years ago
HELP!!! PLEASE!!!
elena-14-01-66 [18.8K]
<span>Molarity is expressed as the number of moles of solute per volume of the solution. We solve the problem above as follows:

0.1000 mol Mg(NO3)2 / L (.1 L) ( </span><span>148.3 g / mol ) = 1.483 g Mg(NO3)2

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Hope this answers the question. Have a nice day.
8 0
3 years ago
A gas has a pressure of 3 atm at 350°C. What will its pressure be at 250°C? The volume and amount of gas is constant. Hint: Conv
vichka [17]

<u>We are given:</u>

P1 = 3 atm                  T1 = 623 K <em>(350 + 273)</em>

P2 = x atm                 T2 = 523 K <em>(250 + 273)</em>

<em />

<u>Solving for x:</u>

From the idea gas equation:

PV = nRT

since number of moles (n) , Volume (V) and the Universal Gas constant(R) are constants;

P / T = k   (where k is a constant)

the value of  k will be the same for a gas with variable pressure and temperature and constant moles and volume

Hence, we can say that:

P1 / T1 = P2 / T2

3 / 623 = x / 523

x = 523 * 3 / 623

x = 2.5 atm (approx)

Therefore, the final pressure is 2.5 atm

7 0
2 years ago
How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

4 0
3 years ago
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