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Free_Kalibri [48]
3 years ago
15

The specific heat of liquid water is 4.184 j/g· ?c. calculate the energy required to heat 10.0 g of water from 26.5?c to 83.7?c.

1. 572 j 2. 837 j 3. 2.39 × 103 j 4. none of these 5. 239 j
Chemistry
1 answer:
Roman55 [17]3 years ago
5 0
Energy required=mass*specific heat*temperature change
=10*4.184*57.2
=2393.248j
=2.39*10^3
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Answer:

270

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hope this helps!

7 0
3 years ago
An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +
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Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

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The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

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\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

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7 0
3 years ago
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Gre4nikov [31]

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Explanation:

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(In attachment)

Formal\, charge = Valence\, electron - (nonbonding\, electrons + \frac{Bonding\,electrons}{2})

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6 0
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