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3 years ago

How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?

Chemistry

1 answer:

BartSMP [9]3 years ago

7 0

Answer:

$m_{AgNO_3}=577.6mg$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:

$n_{LiOH}=81.5mg*\frac{1g}{1000mg}*\frac{1mol}{23.95g} =0.0034molLiOH$

Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:

$m_{ AgNO_3}=0.0034molLiOH*\frac{1molAgNO_3}{1molLiOH} *\frac{169.87gAgNO_3}{1molAgNO_3} *\frac{1000gAgNO_3}{1gAgNO_3} \\\\m_{AgNO_3}=577.6mg$

Best regards!

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3 years ago

Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

Answer:

For 1: The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2: The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

Explanation:

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

For 1:

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2:

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

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