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3 years ago

10

What is likely to happen to the population of thrushes the population of voles decreases?

1 answer:

Andrei [34K]3 years ago

7 0

Answer:

I think it will decrease

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How does heat, without causing melting, damage a rock below eath’s surface

zheka24 [161]

It could be erosion

4 0

3 years ago

A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ris

Blababa [14]

<u>Answer:</u> The heat capacity of calorimeter is $15.66J/^oC$

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

$\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released = ?

n = number of moles = 0.473 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -1235 kJ/mol = $-1235\times 10^3J/mol$ (Conversion factor: 1 kJ = 1000 J)

Putting values in above equation, we get:

$-1235\times 10^3J/mol=\frac{q}{0.473mol}\\\\q=(-1235\times 10^3J/mol\times 0.473mol)=-584.16\times 10^3J$

To calculate the heat capacity of calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed by the calorimeter = $584.16\times 10^3J$

c = heat capacity = ?

$\Delta T$ = change in temperature = $T_2-T_1=62.3^oC-25^oC=37.3^oC$

Putting values in above equation, we get:

$584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC$

Hence, the heat capacity of calorimeter is $15.66J/^oC$

4 0

4 years ago

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the

Sergio [31]

Answer:

$1.2\times 10^{-2}$

Explanation:

The given reactions are:

PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq) $K_3 = 1.89\times 10^{-10}$

AgCl(aq)⇌Ag+(aq)+Cl−(aq) $K_4 = 1.25\times 10^{-4}$

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

$K_f = \frac{K_3}{K_4^2}\\=\frac{1.89\times 10^{-10}}{1.25\times 10^{-4}}^2\\=1.2\times 10^{-2}$

4 0

4 years ago

In the lab, frank has two solutions that contain alcohol and is mixing them with each other. he uses 100 milliliters less of sol

VladimirAG [237]

Solution B (17% alcohol) = x ml
Solution A (13% alcohol) = (x-100) ml

0.17x ml alcohol in Solution B
0.13(x-100) ml alcohol in Solution A

0.17x + 0.13(x-100) = 347
0.17x+0.13x-13=347
0.3x = 360
x=360/0.3=3600/3=1200 ml solution B

7 0

3 years ago

Read 2 more answers

At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.

natka813 [3]

Explanation:

The given data is as follows.

T = 298 K, $\Delta H^{o}$ = -5645 kJ/mol

$\Delta G^{o}$ = -5798 kJ/mol

Relation between $\Delta H$ and $\Delta G$ are as follows.

$\Delta G^{o}$ = $\Delta H^{o} - T \Delta S^{o}$

-5798 kJ/mol = -5645 kJ/mol - $298 \times \Delta S^{o}$

-153 kJ/mol = - $298 \times \Delta S^{o}$

$\Delta S^{o}$ = 0.513 kJ/mol K

Now, temperature is $37^{o}C$ = (37 + 273) K = 310 K

Since, $\Delta G$ = $\Delta H^{o} - T \Delta S^{o}$

= $-5645 kJ/mol - 310 K \times 0.513 kJ/mol K$

= (-5645 kJ/mol - 159.03 kJ/mol)

= -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

$\Delta G = \Delta G_{310 K} - \Delta G_{298 K}$

= -5804.03 kJ/mol - (-5798 kJ/mol)

= -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0

3 years ago