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Amiraneli [1.4K]
3 years ago
5

Identify the structure that protects and gives support to this plant cell.

Chemistry
2 answers:
erik [133]3 years ago
8 0
Cytoskeleton, because skeletons support structures
padilas [110]3 years ago
6 0
Cell wall
Giddy Up!!!!!!!!!!!
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A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:
soldi70 [24.7K]

Answer:

2.01V ( To three significant digits)

Explanation:

First we show the standard reduction potentials of Cu2+(aq)/Cu(s) system and Al3+(aq)/Al(s) system. We can clearly see from the balanced redox reaction equation that aluminium is the anode and was the oxidized specie while copper is the cathode and was the reduced specie. This observation is necessary when substituting values of concentration into the Nernst equation.

The next thing to do is to obtain the standard cell potential as shown in the image attached and subsequently substitute values of concentration and standard cell potential into the Nernst equation as shown. This gives the cell potential under the given conditions.

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3 years ago
Can someone help, please ??
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7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
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