((3.4844 x 10^(-4)  (°C)/hpa (999))+58.3% (-2.52 x 10^(-6)  g/ml*22.8)+2.0582 x10 ^(-5)  °C)/(22.8°C+273.15°C)

Let's break down and solve the given expression step by step:

\[
\frac{\left( (3.4844 \times 10^{-4} \, \text{°C/hPa} \times 999) + 58.3\% \times (-2.52 \times 10^{-6} \, \text{g/ml} \times 22.8) + 2.0582 \times 10^{-5} \, \text{°C} \right)}{22.8 \, \text{°C} + 273.15 \, \text{°C}}
\]

### Step 1: Calculate each part within the numerator
1. \(3.4844 \times 10^{-4} \, \text{°C/hPa} \times 999\)
   \[
   3.4844 \times 10^{-4} \times 999 = 0.348056 \, \text{°C/hPa}
   \]

2. \(58.3\% \times (-2.52 \times 10^{-6} \, \text{g/ml} \times 22.8)\)
   \[
   58.3\% = 0.583
   \]
   \[
   -2.52 \times 10^{-6} \times 22.8 = -5.7456 \times 10^{-5}
   \]
   \[
   0.583 \times (-5.7456 \times 10^{-5}) = -3.3475248 \times 10^{-5}
   \]

3. \(2.0582 \times 10^{-5} \, \text{°C}\)
   This value remains the same: \(2.0582 \times 10^{-5} \, \text{°C}\).

### Step 2: Sum the values in the numerator
\[
0.348056 + (-3.3475248 \times 10^{-5}) + 2.0582 \times 10^{-5} = 0.348056 - 1.2893248 \times 10^{-5}
\]
\[
= 0.348043106752
\]

### Step 3: Calculate the denominator
\[
22.8 \, \text{°C} + 273.15 \, \text{°C} = 295.95 \, \text{°C}
\]

### Step 4: Final division
\[
\frac{0.348043106752}{295.95} = 0.00117613
\]

### Final Answer:
The value of the given expression is approximately \(0.001176 \, \text{°C/hPa}\).

Would you like further details or have any questions?

Here are some related questions for further practice:
1. How do you handle scientific notation in calculations?
2. How does pressure affect temperature in meteorological equations?
3. What are the units of the final result, and why?
4. How do percentage factors change the results in mixed equations?
5. What role does the constant 273.15 °C play in temperature conversions?

**Tip**: Always ensure that the units in a formula are consistent to avoid errors in the final result.

Explore how to solve an expression involving scientific notation, unit conversions, and percentage-based calculations. This problem demonstrates how to break down complex expressions step by step, calculate precise results, and maintain unit consistency. Suitable for Grades 9-12, this example provides a hands-on understanding of physical equations.

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