# A substance has a density of 0.83 g/mL. Which of the following is true?- The substance will SINK because it’s density

A substance has a density of 0.83 g/mL. Which of the following is true? - The substance will SINK because it's density is GREATER than water
- The substance will SINK because it's density is SMALLER than water
- The substance will FLOAT because it's density is SMALLER than water
- The substance will FLOAT because it's density is GREATER than water

## This Post Has 10 Comments

1. Chris3529 says:

it's b

Explanation:

2. benjamingom030 says:

C.Yes. The density equals that of pure silver.

Explanation:

Yes the unknown metal is pure silver.

Density is the ratio of the mass of the substance to its volume. Density is directly proportional to that if mass and inversely proportional to that of volume.

Density = $\frac{mass }{ volume}$ = $\frac{672 g }{64 cm3}$ cm³

= 10.5 g/cm³

Here the density of the unknown metal is the same as that of silver.

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4. qholmes02 says:

omparing the values ​​of the strung density, the one that is closest to carbon is number 2

Explanation:

In this exercise we are given the mass and volume of a body, we are asked to calculate the density, using the equation

ρ = m / V

in the third and fourth column of the table is the density and the substance with the closest value

    mass   volume  density   material

(gr)      (cm³)       (gr/cm³)

1      6.95      4.0          1.74       magnesium

2     4.54      2.0          2.27       carbon

3     5.40      3.0          1.80        phosphorus

4    10.35      5.0         2.07        sulfur

When comparing the values ​​of the strung density, the one that is closest to carbon is number 2

5. kekeke68 says:

Lab 1 - Density Determinations and Various Methods to Measure Volume

The uncertainty in the volume must be determined by error propagation. Mass, length, and diameter measurements contribute to the overall uncertainty.

Volume by pycnometry

Pycnometry is a technique that uses the density relationship between volume and mass, and the vessel used is called a pycnometer .

To perform pycnometry measurements, the mass of the cylinder and the mass of a flask filled with water to a mark (A, Fig. 3) are recorded. The cylinder is then inserted into the flask. Water is displaced when the cylinder is inserted. The volume of water displaced is removed by pipet, thereby restoring the water level to the mark (B). The combined mass of the flask, remaining water, and cylinder is then measured.

Figure 3

Figure 3

The sums of the masses before and after are equal. The massA, the massB, and the masscylinder were all measured on the balance. There is only one unknown in the equation - the mass of the displaced water.

( 4a )

massA + masscylinder = massB + massdisplaced water

massdisplaced water = massA + masscylinder − massB

The volume of water removed is equal to the volume of the cylinder. Masswater can be converted to volume using the density of water.

( 4b )

Vdisplacedwater = Vcylinder = massdisplaced water / densitywater

The density of the cylinder is calculated using mcyl/Vcyl.

The uncertainty calculation requires a few steps and assumptions. The volume of the cylinder was equal to the volume of the water. Vwater was based on the three mass measurements - the mass of the cylinder, of A, and of B.

The uncertainty in masscylinder comes from the balance reading.

The uncertainty associated with massA and massB depends on your ability to precisely adjust the level of the water to the mark at the exactly same place every time (calibration). By repeatedly filling the flask to the mark and taking the mass readings, the average mass of A and the standard deviation (the fluctuation in the mass due to variations in the exact liquid level) can be found.

( 4c )

mA

=

mA,trial1 + mA, trial2 + mA, trial3 +

trials

( 4d )

σmA = ±

mA

-mA,trial1

2+

mA

-mA,trial2

2+

# trials-1

Assume the uncertainty in the mass of both A and B is the same: mA ± σmA; mB ± σmA.

The uncertainty in the mass of water displaced is determined by error propagation:

( 4e )

σmwater = σmA + σmB + σmcyl = σmA + σmA + σmcyl

The density of water at room temperature is known quite precisely and is assumed to contribute negligible error (see table at the end of the lab), so dividing σm, water by the density of water to give σV, water is adequate. Since σV, water = σV, cyl, the uncertainty in the density can be determined.

( 4f )

σρ = ± ρ

σV

V

+

σm

m

You will use pycnometry in parts 4 and 5 to determine the volume and/or density of a hollow cylinder and of a mixed cylinder.

Volume of a void inside a hollow cylinder

A hollow cylinder has an empty space inside.

Figure 4

Figure 4

The volume of the cylinder is comprised of the volume of metal and the volume of the void inside.

( 5a )

Vcyl = Vmetal + Vvoid → Vvoid = Vcyl - Vmetal

Vcyl is determined by pyncometry. The volume occupied by the metal can be determined using the mass of the cylinder (which is due to only the metal, not the void) and the density of the metal, which was determined previously in the lab (either Al or brass, depending on the cylinder). Use the value for density that is closest to the literature values - 2.70 g/cm3 for Al; between 8 and 9 g/cm3 for brass.

( 5b )

Vmetal =

mcyl

ρmetal

No error propagation is required

Percent composition of a mixed cylinder

The total mass of the cylinder, mcyl, is the sum of the mass of Al and brass (mAl + mbrass). In terms of fractional composition, this would be Xmcyl and (1 - X)mcyl, respectively, where X is the Al fraction and (1-X) is the brass fraction (the remainder).

The cylinder volume is determined by pycnometry and is the sum of the volumes of the two metals:

( 6a )

Vcyl = VAl + Vbrass

Replace each volume by its mass divided by its density using V=m/ρ:

( 6b )

Vcyl =

mAl

ρAl

+

mbrass

ρmass

Replace the masses by the equivalent expressions in terms of X and mcyl:

( 6c )

Vcyl =

X mcyl

ρAl

+

(1−X)mcyl

ρbrass

Divide through by mcyl and replace Vcyl/mcyl with 1/ρcyl:

( 6d )

1

ρcyl

=

X

ρAl

+

(1 − X)

ρbrass

Collect terms on the right-hand side that contain X:

( 6e )

1

ρcyl

= X

1

ρAl

1

ρbrass

+

1

ρbrass

6. sarahcyberpony says:

1. True

2. False

4. True

4. True

Explanation:

2. warm liquid is less dense and therefore will rise. while cool liquid is more dense and therefore will sink

7. aronhaile says:

the awnser is A

Explanation:

Edge 2020

8. mexprencss says:

Option (c) is correct.

Explanation:

The density of a substance is 0.83 g/mL.

We know that the density of water is 1 g/mL

It means the density of the substance is less that that of density of water. If the density of substance is less that water then it will float on water's surface.

Hence, the correct option is (c) "The substance will FLOAT because it's density is SMALLER than water".

9. chayaharroch03 says:

C. Yes, the density equals that of pure silver (10.5)

Step-by-step explanation: