Welcome to the article Calculate the volume of a tex 225 g tex liquid with a density of tex 1 39 g mL tex A 162 L B 313. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!
Answer :
To find the volume of a liquid, we can use the formula:
[tex]\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \][/tex]
Here are the given values:
- Mass ([tex]\(m\)[/tex]) = 225 grams
- Density ([tex]\(\rho\)[/tex]) = 1.39 grams per milliliter (g/mL)
Now, let's plug these values into the formula and solve for the volume:
[tex]\[ \text{Volume} = \frac{225 \, \text{g}}{1.39 \, \text{g/mL}} \][/tex]
When we calculate this, we get:
[tex]\[ \text{Volume} \approx 161.87 \, \text{mL} \][/tex]
So, the volume of the 225 grams of liquid with a density of 1.39 grams per milliliter is approximately [tex]\(161.87 \, \text{mL}\)[/tex].
Out of the given options:
- 162 L
- 313 mL
- 162 mL
- 313 L
The option that best matches the calculated volume is:
[tex]\[ \boxed{162 \, \text{mL}} \][/tex]
[tex]\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \][/tex]
Here are the given values:
- Mass ([tex]\(m\)[/tex]) = 225 grams
- Density ([tex]\(\rho\)[/tex]) = 1.39 grams per milliliter (g/mL)
Now, let's plug these values into the formula and solve for the volume:
[tex]\[ \text{Volume} = \frac{225 \, \text{g}}{1.39 \, \text{g/mL}} \][/tex]
When we calculate this, we get:
[tex]\[ \text{Volume} \approx 161.87 \, \text{mL} \][/tex]
So, the volume of the 225 grams of liquid with a density of 1.39 grams per milliliter is approximately [tex]\(161.87 \, \text{mL}\)[/tex].
Out of the given options:
- 162 L
- 313 mL
- 162 mL
- 313 L
The option that best matches the calculated volume is:
[tex]\[ \boxed{162 \, \text{mL}} \][/tex]
Thank you for reading the article Calculate the volume of a tex 225 g tex liquid with a density of tex 1 39 g mL tex A 162 L B 313. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Brahmana