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aliya0001 [1]
3 years ago
12

Would you measure the height of a building in meters?

Chemistry
2 answers:
olga nikolaevna [1]3 years ago
6 0
<span>Yes. If you are using the metric system, then this the most accurate </span>
sweet-ann [11.9K]3 years ago
3 0
Yes but really it depends on how tall the building is
You might be interested in
Give an example of how you could use different reference frames for the same motion.
RUDIKE [14]

Answer:

Motion is defined as a change of position. How we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion.

Explanation:

8 0
2 years ago
Read 2 more answers
What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same t
kenny6666 [7]

Answer: 20.7 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system=-P\Delta V  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = 4.51\times 4.184kJ=18.9kJ    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

\Delta E=+1.79+(18.9)=20.7kJ

Thus \Delta E for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ

4 0
2 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
1 year ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
2 years ago
The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o
belka [17]

Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

8 0
3 years ago
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