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Pavlova-9 [17]
3 years ago
15

What do you think could be happening to the sand eels​

Chemistry
2 answers:
Alex777 [14]3 years ago
7 0

Answer:

is there like a passage to answer this question?

Explanation:

kogti [31]3 years ago
7 0

Answer:

Increasing fishing for them is thought to be causing problems for some of their natural predators, especially the auks which take them in deeper water. They are also tied as flies to catch fish. An instance of this was the RSPB report linking a population crash of seabirds in the North Sea to fishing for sand eels.

Explanation:

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What kind of energy can be transferred?<br><br> Help quick please.
defon

Answer:

Any kind, as long as there is an action.

7 0
3 years ago
Read 2 more answers
Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2
pashok25 [27]

Answer : The value of \Delta H for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)     \Delta H_1=-571.6kJ/mole

Now we have to determine the value of \Delta H for the following reaction i.e,

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)    \Delta H_2=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=-(-571.6kJ/mole)

\Delta H_2=+571.6kJ/mole

Hence, the value of \Delta H for the reaction is +571.6 kJ/mole.

6 0
3 years ago
Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
2 years ago
what is the main difference between organisms that share many characteristics and organisms that do not
Anarel [89]
Organisms that share many derived characteristics are probably more closely related
8 0
3 years ago
For the reaction below, complete the rate expression that relates the change in concentration with respect to time to the rate o
Ann [662]

Answer: Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of Cl_2= -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = \frac{1d[NOCl]}{2dt}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO+Cl_2\rightarrow 2NOCl

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of  = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of = -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = +\frac{1d[NOCl]}{2dt}

5 0
2 years ago
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