Why are the chemicals in the water turning the phrogs gay?

Give reason nitrogen is a gas Help:)

Mrrafil [7]

Nitrogen is a gas because it is an airlike fluid substance which expands freely to fill any space available, irrespective of it's quantity

7 0

4 years ago

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A quality of gas under a pressure of 600.0mm Hg has a volume of 39.7 liters. What is it’s volume at standard pressure when the t

Leokris [45]

Answer: it would be 23820 liters i dont know thats my best answer, hope it helps!

Explanation:

you have to multiply it or divide it.

Hope it helps!! :)

3 0

3 years ago

Which part of Earth most likely has the highest albedo value?

andre [41]

Uhhh pretty sure B

Happy New Years

3 0

3 years ago

Calculate the number of hydrogen atoms in a sample of hydrazine . Be sure your answer has a unit symbol if necessary, and round

gladu [14]

Answer:

$atoms \ H= 9.767x10^{24}atoms$

Explanation:

Hello!

In this case, considering that the mass of hydrazine is missing, we can assume it is 130.0 g (a problem found on ethernet). In such a way, since we need a mass-mole-atoms relationship by which we can compute moles of hydrazine given its molar mass (32.06 g/mol), then the moles of hydrogen considering one mole of hydrazine has four moles of hydrogen and one mole of hydrogen has 6.022x10²³ atoms (Avogadro's number); therefore, we proceed as shown below:

$atoms \ H=130.0gN_2H_4*\frac{1molN_2H_4}{32.06gN_2H_4} *\frac{4molH}{1molN_2H_4} *\frac{6.022x10^{23}atoms}{1molH}\\\\atoms \ H= 9.767x10^{24}atoms$

Notice 130.0 g has four significant figures, therefore the result is displayed with four as well.

Best regards!

7 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

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

where,

$\Delta G^o$ = standard Gibbs free energy = ?

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

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago