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Lemur [1.5K]
3 years ago
6

One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal

Chemistry
1 answer:
kenny6666 [7]3 years ago
8 0

Answer:

Lithium oxide, Li₂O.  

Explanation:

Hello!  

In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:

M_2O+H_2\rightarrow 2M+H_2O  

Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:

5.00gM_2O*\frac{1molM_2O}{(2*X+16)gM_2O}*\frac{2molM}{1molM_2O}*\frac{XgM}{1molM}=2.32gM  

Thus, we perform the operations in order to obtain:

\frac{10X}{(2X+16)}=2.32  

So we solve for x as shown below:

10X=2.32(2X+16)\\\\10X=4.64X+37.12\\\\X=\frac{37.12}{10-4.64}\\\\X= 6.93g/mol  

Whose molar mass corresponds to lithium, and therefore, the metallic oxide is lithium oxide, Li₂O.  

Best regards!

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\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol

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\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol

For , 2.41 moles of I_2 :

\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ

We know :

\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K

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Explanation:

Bernoulli equation for the flow between bottom of the tank and pipe exit point is as follows.

   \frac{p_{1}}{\gamma} + \frac{V^{2}_{1}}{2g} + z_{1} = \frac{p_{2}}{\gamma} + \frac{V^{2}_{2}}{2g} + z_{2} + h_{f} + h_{t}

    \frac{(100 \times 144)}{62.43} + 0 + h[tex] = [tex]\frac{(50 \times 144)}{(62.43)} + \frac{(70)^{2}}{2(32.2)} + 0 + 40 + 60

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                                = 52.684%

Thus, we can conclude that the efficiency of the turbine is 52.684%.

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