My teacher laid this much out for us but I don’t know how to get the products in each one.

The solvent is usually referred to as the component of a solution which is present as

Eddi Din [679]

The solvent is usually referred to as the component of a solution which is present as the one with the larger quantity and in most cases as the liquid which dissolves a solid. In a solution, there are two components namely the solvent and the solute. The solute is the one in smaller amount.

8 0

3 years ago

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

Which of the following describes a strong acid?

zlopas [31]

I remember learning this last y’all i jus don’t remember it might be A or C

7 0

3 years ago

Which is the correct Lewis structure for carbon monoxide?

LuckyWell [14K]

<span>The Lewis structure for CO has 10 valence electrons. For the CO Lewis structure you'll need a triple bond between the Carbon and Oxygen atoms in order to satisfy the octets of each atom while still using the 10 valence electrons available for the CO molecule.</span>

7 0

3 years ago

Carbon 14 decays to Carbon 12 and has a half-life of 5730 years. If a fossil is analysed and it has 5 grams of Carbon 14 and 15

Katen [24]

Answer:

im sorry but this question doesnt make sense

Explanation:

7 0

2 years ago