Q: What is a volume of 5.0 g pure water?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

The density of pure water at +4 degrees C, is 1 g/ml. The volume, then, of 5000 kg of water is 5000 liters.

Foam.

Assuming the gold nugget is solid and pure (unlikely in real life!) The volume of the nugget is 77.0 - 50.0 mL = 27.0 mL So density = mass/volume = 521/27.0 = 19.3 g per mL.

Just divide the mass by the volume.

50/2.6 = 19.231 gm/cm3 (rounded)

Related questions

The density of pure water at +4 degrees C, is 1 g/ml. The volume, then, of 5000 kg of water is 5000 liters.

Given that the volume of water displaced by the body is 50 mL (from 150 mL to 200 mL), and since 1 mL of water is equivalent to 1 g, the mass of the body is equal to the volume of water displaced, which is 50 g. Therefore, the density of the body is 50 g / 60 g = 0.83 g/mL.

The volume of 250g of sea water will be slightly less than the volume of the same mass of pure water due to the presence of dissolved salts and minerals in sea water. The dissolved solids in sea water increase its density, thus requiring less volume to accommodate the same mass compared to pure water.

The mass of 0.23 kg of pure water is equal to 230 grams. Given that the density of pure water is 1 g/ml, the volume of 230 grams of water would be 230 ml.

The density of water at room temperature is approximately 1 g/mL. So, for a 50 mL sample that weighs 98.5 g, the density can be calculated as mass/volume = 98.5 g / 50 mL = 1.97 g/mL.

The volume of 35.7 grams of water = 35.7 cubic centimetres at standard temperature and pressure, (STP). This means a sample at 0°C at a pressure of one atmosphere.

The volume of the coin can be calculated using its density, which is approximately 0.379 cm^3/g for pure silver. With a mass of 16.0 g, the volume of the coin would be 16.0 g / 0.379 cm^3/g = 42.2 cm^3.

Foam.

At "room temperature" (20 deg Celsius" and at a pressure of 1 atmosphere, the density of water is 998.2071 kilograms per cubic metre = 0.998 207 1 grams per millilitre.So the mass of 50 millilitres of pure idle, under those conditions, is 49.91 grams, not 50 grams as some might claim.Pure water attains its maximum density, of 999.9720 kg/m3at 4 degrees Celsius. At normal pressure, the densityneverreaches the value 1.

The **density** of a substance is defined as its mass per unit volume. We can calculate the density using the formula: [ \text{Density} (\rho) = \frac{\text{Mass} (m)}{\text{Volume} (V)} ] Given that the sample has a volume of **50 cm³** and a mass of **135 g**, let's determine the density: [ \rho = \frac{135 , \text{g}}{50 , \text{cm³}} ] The calculated density is approximately **2.7 g/cm³**[^10^]. Now let's compare this value to known densities: **Gold**: Gold has a density of *19.3 g/cm³*⁷. The sample's density is significantly lower. **Pure Water**: The density of pure water is approximately **1 g/cm³** at 4.0°C (39.2°F) . The sample's density is higher than water. **Aluminum**: Aluminum has a density of *2.7 g/cm³*[^10^]. The sample's density matches that of aluminum. **Ocean Water**: Ocean water contains dissolved salts, which increase its density. Seawater density typically ranges from *1.02 g/cm³ to 1.03 g/cm³*. The sample's density is higher than seawater. Based on the calculated density, the sample is most likely **aluminum**.

Since the density (mass divided by volume) of water is about 1.0 g/mL the volume of 6.5 gram is6.5(g) / 1.0(g/mL) = 6.5 mL

The density of the substance is calculated by dividing the mass (50 g) by the volume (75 mL). Density = mass/volume. Therefore, the density of the substance is 0.67 g/mL.