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3 years ago

An atom of helium-4 differs from an atom of lithium-7 in that the atom of helium-4 has?

Chemistry

1 answer:

lianna [129]3 years ago

4 0

2.valence electrons while lithium has 5

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Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3 ,what is the samples volume?

kirill115 [55]

Answer: 10 cm3

Explanation:

Mass of sample of steel (m) = 80 grams

density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?

Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:

density = mass / volume

p = m / v

make v the subject formula

v = m / p

v = 80 grams / 8 g/cm3

v = 10 cm3

Thus, the samples volume is 10 cm3

4 0

2 years ago

The nucleus contains both protons and electrons True or false

Varvara68 [4.7K]

Answer: False

Explanation: The nucleus of an atom only contains the protons and neutrons.

The electrons are not found in the nucleus,

they are orbiting the nucleus in different shells.

3 0

3 years ago

Is the process by which certain waste materials, such as plastic, glass, aluminum, and paper, are prepared for reuse.

andriy [413]

the process is called recycling.

3 0

3 years ago

Read 2 more answers

How many moles are there in 24.0 grams of H2O

Nata [24]

Answer:So, one mole of water has a mass of 16 +1+1 = 18 grams. So, if one mole has a mass of 18 grams, 25 grams would have a mass of 25 grams/ 18 grams per mole or 1.39 moles

6 0

2 years ago

Read 2 more answers

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

Explanation:

Wavelength of the photon emitted = $\lambda =646.3 nm =646.3\times 10^{-9} m$

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

$E=\frac{hc}{\lambda }$

h = Planck's constant = $6.626\tiomes 10^{-34} Js$

c = Speed of the light = $3\times 10^8 m/s$

$E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}$

$E=3.07\times 10^{-19} J$

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

3 0

3 years ago