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Zolol [24]
3 years ago
10

Which of the following is not the same as 0.00684 kg? 0.0684 dkg,6.84 g, 684 cg, 6,840 mg,

Chemistry
2 answers:
lesya [120]3 years ago
8 0
The Answer would be C.) 684 cg


emmainna [20.7K]3 years ago
6 0

Answer : 0.00684 kg is not equal to 0.0684 dkg.

Explanation :

The conversion used for mass from kilogram to dekagram is :

1 kg = 100 dkg

The conversion used for mass from kilogram to grams is :

1 kg = 1000 g

The conversion used for mass from kilogram to centigrams is :

1 kg = 100000 cg

The conversion used for mass from kilogram to milligrams is :

1 kg = 1000000 mg

Now we have to determine the same mass as 0.00684 kg.

(1) 1 kg = 100 dkg

0.00684 kg = \frac{0.00684kg}{1kg}\times 100dkg=0.684dkg

(2) 1 kg = 1000 g

0.00684 kg = \frac{0.00684kg}{1kg}\times 1000g=6.84g

(3) 1 kg = 100000 cg

0.00684 kg = \frac{0.00684kg}{1kg}\times 100000cg=684cg

(4) 1 kg = 1000000 mg

0.00684 kg = \frac{0.00684kg}{1kg}\times 1000000mg=6840mg

Hence, from this we conclude that, 0.00684 kg is not equal to 0.0684 dkg.

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Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
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<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

4 0
3 years ago
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