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2 years ago

Help me please really only answer if u know the answer

Chemistry

2 answers:

Shtirlitz [24]2 years ago

7 0

Answer:

Explanation:

Asafahs

iren [92.7K]2 years ago

4 0

Answer:

Genetic matierial that is identical.

Explanation:

Because it produces new sets of chromosomes. This is biology btw. But you can never trust anyone on the internet.

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What mass of C6H8O7 should be used every 7.0 X 10^2mg NaHCO3

Savatey [412]

Mass C₆H₈O₇ : 0.531484 g

<h3>Further explanation</h3>

Reaction

3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)

MW NaHCO₃ : 84 g/mol

mass NaHCO₃ : 7.10² mg=0.7 g

mol NaHCO₃ :

$\tt mol=\dfrac{0.7}{84}=0.0083$

mol C₆H₈O₇ :

$\tt \dfrac{1}{3}\times 0.0083=0.00277$

MW C₆H₈O₇ : 192 g/mol

mass C₆H₈O₇ :

$\tt mass=0.00277\times 192=0.53184$

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2 years ago

Hypothesize How much do you think an apple weighs?

umka2103 [35]

Answer: 3 pounds

Explanation:

8 0

2 years ago

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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

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2 years ago

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Mazyrski [523]

Answer:

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What aspect of metallic bonding is responsible for malleability ductility and conductivity of metals?

ANTONII [103]

Bonds are forces of attractions between atoms formed by the transfer of electrons or sharing of electrons. Metallic bond is a type bond that exist in metallic structures where the atoms of the metals attracts the sea of electrons in the structure.It is these metallic bonds that results to the malleability , ductility and conductivity of metals because in that the sea of electrons makes them conduct electricity. In addition the atoms of metals in the structure are ions which can slide past each other in the sea of electrons.

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2 years ago