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4 years ago

What is adaptation, and what are some examples of it?

1 answer:

6 0

An adaption is a change or process an organism goes through to become better suited for its environment.

Example: A polar bear having white fur in order to blend in in its environment.

You might be interested in

When the concentration of SO2(g) is increased to 1.48 M, the ratio of products to reactants is 1.4. The equilibrium constant for

weeeeeb [17]

Answer:

Towards the products

Explanation:

3 0

3 years ago

What is the net cell reaction for the cobalt-silver voltaic cell? express your answer as a chemical equation?

slava [35]

Answer: The net ionic equation for the cobalt-silver voltaic cell is $Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)$

Explanation: In cobalt-silver voltaic cell, one half of the cell consists of cobalt electrode immersed in $Co(NO_3)_3$ solution ( which means that $Co^{3+}$ are present in the solution) and other half of the cell consists of the Ag electrode immersed in $AgNO_3$ solution ( which means that $Ag^+$ is present in the solution)

The two electrodes are joined by the copper wire. The cobalt electrode acts as an anode and the silver electrode acts a cathode.

At anode, oxidation reaction takes place and at cathode, reduction reaction takes place.

At Anode : $Co(s)\rightarrow Co^{3+}(aq.)+3e^-$

At Cathode: $[Ag^+(aq.)+e^-\rightarrow Ag(s)]\times 3$

Net ionic equation: $Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)$

6 0

3 years ago

Calculate the percent ionization of a 0.18 M benzoic acid solution in a solution containing 0.10 M sodium benzoate.

Yuliya22 [10]

Answer:

% I = 0.083 %

Explanation:

C6H5COOH + NaOH ↔ NaC6H5CO2 + H2O

∴ M C6H5COOH = 0.18 mol/L

∴ M NaC6H5CO2 = 0.10 mo/L

% ion = ( { H3O+ ] / initial acid concentration ) * 100

⇒ C6H5COOH ↔ H3O+ + C6H5COO-

⇒ NaC6H5CO2 ↔ Na+ + C6H5COO-

∴ Ka = 6.4 E-5 = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ]

Ka value taken from the literature

mass balance

⇒ 0.18 + 0.1 = [ C6H5COOH ] + [ C6H5COO- ]

⇒ [ C6H5COOH ] = 0.28 - [ C6H5COO- ] ........(1)

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ C6H5COO- ]

∴ [ Na+ ] ≅ M NaC6H5CO2 = 0.1 M

⇒ [ C6H5COO- ] = [ H3O+ ] + 0.1..............(2)

(1) and (2) in Ka:

⇒ 6.4 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.1 ) / ( 0.28 - ( [ H3O+ ] + 0.1 ) )

⇒ 6.4 E-5 = [ H3O+ ]² + 0.1 [H3O+ ] / ( 0.18 - [ H3O+ ] )

⇒ 1.17 E-5 - 6.5 E-5 [ H3O+ ] = [ H3O+ ]² + 0.1 [ H3O+ ]

⇒ [ H3O+ ]² + 0.1 [ H3O+ ] - 1.17 E-5 = 0

⇒ [ H3O+ ] = 1.493 E-4 M

⇒ % I = ( 1.493 E-4 / 0.18 ) * 100 = 0.083 %

6 0

3 years ago

A solution is made by dissolving 4.35 g of glucose (C6H1206) in 25.0 mL of water at 25 °C.Calculate the molality of glucose in t

Roman55 [17]

Answer:

The molality is unchanged (0.96 molal)

Explanation:

Step 1: Data given

mass of glucose = 4.35 grams

volume of water = 25.0 mL

Density of water = 1.00 g/mL

Molar mass of glucose = 180.156 g/mol

Step 2: Calculate number of moles

moles of glucose = mass of glucose / Molar mass of glucose

moles of glucose = 4.35 grams / 180.156 g/mol

moles of glucose = 0.024 moles

Step 3: Calculate mass of water

mass = density * volume

mass of water = 1.00 g/mL * 25.0 mL

mass of water = 25 g = 0.025 kg

Step 4: Calculate molality

molality = Number of moles / mass of water

molality = 0.024 moles / 0.025 kg

molality = 0.96 molal

Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled.

This means double mass of water = 2*0.025 kg = 0.050 kg

Now molality is 0.024 moles / 0.050 kg = 0.48 molal

When the mass of solvent is doubled, the molality is halved from 0.96 molal to 0.48 molal

You take this newsolution and add more solute, so that the original mass of the solute is doubled.

This means double mass of glucose = 2*4.35 g = 8.70 g

8.70 grams of glucose = 8.7 grams * 180.156 g/mol = 0.048 moles

molality = 0.048 moles / 0.050 kg = 0.96 molal

The molality is unchanged

5 0

4 years ago

Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

Answer: The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

For process 2:

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

For process 3:

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

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

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

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

8 0

3 years ago