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3 years ago

Please answer soon!!!!!

Chemistry

1 answer:

fgiga [73]3 years ago

5 0

Answer:

7.27 grams

Explanation:

Format= starting amount (g)·(1 mol/molar mass)·(amount of wanted/amount of starting)·(molar mass of wanted/1 mol)

$41.0 x \frac{1 mol}{17.031}x\frac{3 H2}{2NH3} x\frac{2.016}{1 mol} =7.27$

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PLEASE HELPPPPP!!!!!!!

marin [14]

Boiling water is a physical change, because it is changing state of matter, there is a controllable temperature change, and you can change it back.

Frying the egg white is a chemical change, because there is a change of matter, controllable temperature change, but you CAN'T change it back.

8 0

4 years ago

The equation, DF + heat Imported Asset D + F, is an example of an endothermic reaction.

zvonat [6]

The answer is TRUE.

If the Energy is on the left, then the problem is true. If it is on the right then it would be negative, false, and considered as exothermic.

Endothermic reaction = the products are higher in energy than the reactants.
Exothermic reaction = a chemical reaction that releases energy by light or heat.

4 0

3 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Active

klio [65]

Option B

Primary productivity is not limited by time

<u>Explanation:</u>

To ecosystems, the productivity of the primary producers is essential because certain bodies generate energy for different existence bodies. Besides the era, primary production ends in the extension of dissimilar plant biomass to the practice. Consumers acquire their strength from primary producers, unless directly (herbivores, any detritivores), or diffusely.

It depends superimposed on the availability of the sunlight, availability of the nutrients such as nitrogen, iron, phosphorus from the soil and water. The primary productivity of the plant's distinct autotrophs is not restricted by time.

5 0

4 years ago

Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Misha Larkins [42]

Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

$moles HCl = moles KOH \\ \\ cHCl*vHCl=cKOH*vKOH \\ \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}= 541,54mL$

So, the required volume of HCl is 541,54 mL

Have a nice day!

8 0

3 years ago