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4 years ago

Find the square of (a+b) please answer with method

Chemistry

2 answers:

nataly862011 [7]4 years ago

5 0

(A^2+b^2)

Hope this helps

olga2289 [7]4 years ago

3 0

Answer:

${a}^{2} + 2ab \: + {b}^{2}$

Explanation:

${(a + b)}^{2}$

$(a + b)(a + b)$

${a}^{2} + ab \: + \: ab \: + {b}^{2}$

${a}^{2} + 2ab \: + {b}^{2}$

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Write the half reactions for 2Li + H2 -> 2LiH.

Ksivusya [100]

Li ----> Li+ + e-(oxidation)
now comes 
Reduction: 

H2 + 2e- ------> 2H
now by
Multiplying (1) by 2 and adding in (2) will give you the required equation..
..

2Li + H2 ---------> 2LiH
half reaction are either oxidation or reduction

7 0

3 years ago

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

hichkok12 [17]

Answer:

$K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}$

Explanation:

Hello there!

In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:

$K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}$

Now, when we combine them to get the overall expression, we infer these two are multiplied to get:

$K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}$

Regards!

6 0

3 years ago

PLEASE HELP

WITCHER [35]

The right answer is high number of offspring produced. asexual reproduction only requires on orgainism and creates an identical copy leaving no room for an increase of intelligence. plus it creates much more offspring and more often than sexual reproduction.

4 0

3 years ago

The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of th

kotykmax [81]

Answer:

$\Delta _{fus}H=3255.3J/mol$

$\Delta _{fus}S=7.62\frac{J}{mol*K}$

Explanation:

Hello,

Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

$\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}$

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

$p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )$

Solving for the enthalpy of fusion we obtain:

$\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol$