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Lisa [10]
2 years ago
13

12500 J/ (106 g) (4C) reduce to one

Chemistry
2 answers:
Sloan [31]2 years ago
5 0

The given expression is \frac{12500 J}{(106g)(4^{0}C )}

This expression denotes the specific heat of a substance which indicates a heat energy of 12500 J is involved in raising the temperature of 106 g of the substance by 4^{0}C. Generally, the units of specific heat are \frac{J}{g.^{0}C }

\frac{12500 J}{(106g)(4^{0}C )} = \frac{12500 J}{(106g)(4^{0}C } = 29.5 \frac{J}{g^{0}C }

Therefore,  \frac{12500 J}{(106g)(4^{0}C )} when reduced to one unit will be 29.5 \frac{J}{g^{0}C }

amid [387]2 years ago
5 0

Answer:

29.48 J/g°C

Explanation:

Step 1:

Data obtained from the question. This includes:

12500 J/ (106g) (4°C)

From the question given, we were asked to reduce the expression above to a single value

Step 2:

Reduction of the expression.

The reduction of 12500 J/ (106g) (4°C) can be obtained as follow:

12500 J/ (106g) (4°C) = 12500J/424g°C

= 29.48 J/g°C

Therefore, the reduction of 12500 J/ (106g) (4°C) to a single value is 29.48 J/g°C

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