All categories

3 years ago

Density = _____ (A)weight/length (B)mass/weight (C)mass/volume (D)volume/weight

Chemistry

2 answers:

8_murik_8 [283]3 years ago

7 0

Density= Mass/Volume I am positive I just had an assignment on this

algol133 years ago

5 0

C mass and volume i remember by d=m/v

You might be interested in

Does anyone know the answer to these? Please help.

Phantasy [73]

double replacesment and 1,14,12,6

5 0

3 years ago

bogdanovich [222]

Answer:

2 and 4

Explanation:

4 0

3 years ago

If density is 22.57g/cm^3 what volume would be occupied by 4.80kg of osmium

steposvetlana [31]

Density is equal to the ratio of mass to the volume.

The mathematical expression is given as:

$density=\frac{mass}{volume}$

Arrange the above expression in terms of volume, i.e.

$volume = \frac{mass}{density}$

Density of osmium is equal to = $22.57 g/cm^{3}$

Mass of osmium = 4.80 kg

First, convert the mass in kg to g:

1 kg = 1000 g

Therefore, mass in g = $4.80\times 1000$ = 4800 g

Put the values,

$volume = \frac{4800 g}{22.57 g/cm^{3}}$

= $212.67 cm^{3}$ .

Hence, volume occupied by osmium = $212.67 cm^{3}$ .

4 0

3 years ago

An alpha particle (mass = 6.6 x 10²⁴ g) emitted by radium travels at 3.4 x 10⁷ ± 0.1 x 10⁷ mi/h.

marishachu [46]

Answer:

For a: The De-Broglie wavelength of the given particle is $2\times 10^{-10}m$

For b: The uncertainty in the position is $3.56\times 10^{-14}m$

Explanation:

For a:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{mv}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant =

m = mass of the alpha particle =

v = velocity of alpha particle = $3.4\times 10^7mi/h=1.52\times 10^7m/s$ (Conversion factors used: 1 mile = 1609.34 m & 1 hr = 3600 s)

Putting values in above equation, we get:

$\lambda=\frac{6.6\times 10^{-34}Js}{6.6\times 10^{-27}kg\times 1.52\times 10^7m/s}\\\\\lambda=6.58\times 10^{-15}m$

Hence, the De-Broglie wavelength of the given particle is $2\times 10^{-10}m$

For b:

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p\geq \frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = ?

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of the alpha particle = $6.6\times 10^{-27}kg$

$\Delta v$ = uncertainty in speed = $0.1\times 10^7mi/hr=4.47\times 10^5m/s$

h = Planck's constant = $6.6\times 10^{-34}kgm^2/s$

Putting values in above equation, we get:

$\Delta x=\frac{6.6\times 10^{-34}kgm^2/s}{2\times 3.14 \times 6.6\times 10^{-27}\times kg 4.47\times 10^{5}m/s}\\\\\Delta x=3.56\times 10^{-14}m$

Hence, the uncertainty in the position is $3.56\times 10^{-14}m$

4 0

3 years ago

Which set of values represents standard pressure and standard temperature?(1) 1 atm and 101.3 K(2) 1 kPa and 273 K(3) 101.3 kPa

yuradex [85]

Standard Temperature and Pressure (STP) is 0 degree Celsius and 1 atm. We also know that 1atm= 101.3 kPa. Thus, set 3 is the best choice.
Hope that would help :))

4 0

3 years ago

Read 2 more answers