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2 years ago

. The dissolving process is exothermic when the energy

Chemistry

1 answer:

Serhud [2]2 years ago

8 0

Answer:

Explanation:

The dissolving process depends on the interaction between solute and solvent (solvation) and the breaking up of the intermolecular bond between solutes. The former is exothermic in nature, while the later is endothermic. Energy is released when solute-solvent particles interact. When this energy exceeds the energy required to break intermolecular bonds between the solute particles, dissolution is exothermic.

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Telephone signals are often transmitted over long distances by microwaves. what is the frequency of microwave iradiation with a

Alborosie

<span>c = speed of light = 3.00 x 10^5 km/s = 3.00 x 10^8 m/s
Î» = wavelength of the microwave radiation = 3.50 cm = 0.035 m
f = frequency (in Hertz) = to be determined
f = c/Î» = 3.00 x 10^8 m/s / 0.035 m
f = 8.57 x 10^9 Hz Frequency</span>

6 0

3 years ago

Read 2 more answers

Which substance is the reducing agent in this reaction? 16h++2cr2o72−+c2h5oh→4cr3++11h2o+2co2 express your answer as a chemical

lana66690 [7]

16 H + + 2Cr2O72- + C2H5OH → 4 Cr3 + +11H2O +2CO2

The reducing agent is C2H5OH

Explanation

reducing agent is a substance that loses or donate electrons in a chemical reaction. C2H5OH is the one which donate electrons in the above chemical equation.

7 0

2 years ago

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

If two atoms have the same atomic numbers but different mass numbers, then they can be considered atoms of the same element.

castortr0y [4]

Same elements can have different nuclei. So, true.

7 0

3 years ago

Read 2 more answers

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago