It is 12 to 1 or 12/1 miles of saline to L solution
Answer:
7.23 J
Explanation:
Step 1: Given data
- Mass of graphite (m): 566.0 mg
- Initial temperature: 5.2 °C
- Final temperature: 23.2 °C
- Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹
Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)
Q = 7.23 J
The volume of Muriatic acid needed is 199ml.
<h3>What is concentration?</h3>
- Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.
- Mass concentration, molar concentration, number concentration, and volume concentration are four different categories of mathematical description.
- Any type of chemical mixture can be referred to by the term "concentration," however solutes and solvents in solutions are most usually mentioned.
- There are different types of molar (quantity) concentration, including normal concentration and osmotic concentration.
<h3>How is concentration determined?</h3>
- Subtract the solute's mass from the total volume of the solution. Using m as the solute's mass and V as the total volume of the solution, write out the equation C = m/V.
- To get the concentration of your solution, divide the mass and volume figures you discovered and plug them in.
Learn more about concentration here:
brainly.com/question/13872928
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Explanation:
2. 
First, we need to find the number of moles of
at 300K and 1.5 atm using the ideal gas law:


Now use the molar ratios to find the number of moles of ethane to produce this much
.


Finally, convert this amount to grams using its molar mass:


3. 
Convert 75 g Zn into moles:

Then use the molar ratios to find the amount of H2 produced.

Now use the ideal gas law
to find the volume of H2 produced at 23°C and 4 atm:

