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Kazeer [188]
3 years ago
10

Which is NOT true about oxygen-17 and oxygen-18?

Physics
1 answer:
iren2701 [21]3 years ago
7 0

Answer:

Their atoms have identical mass.

Explanation:

Oxygen-17 and Oxygen-18 are both isotopes of Oxygen.

Oxygen has three isotopes, Oxygen- 16, Oxygen-17 and Oxygen-18.

Oxygen- 16 has 8 protons and 8 neutrons and a mass number of 16.

Oxygen-17 has 8 protons and 9 neutrons and a mass number of 17.

Oxygen-18 has 8 protons and 10 protons and a mass number of 18.

The three isotopes therefore have the same atomic number ( the number of protons) but different mass numbers.

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When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of
Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0
4 years ago
Predict the energy transformation that occurs when carson sands a piece of wood with sandpaper.
tigry1 [53]

Answer:

Explanation:

The hand provides Kinetic Energy in moving.

The KE is transformed to Frictional energy

The Frictional Energy can produce heat and light energy.

The sandpaper produces little shreds from the grit of the paper.

The shreds have KE (they move)

6 0
2 years ago
Which is the only mechanism that can introduce new alleles in a gene pool? genetic drift random mating mutation
Vikentia [17]
Mutation because they change the genes
8 0
3 years ago
1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

6 0
3 years ago
An instrument is defective for all of the following conditions EXCEPT:_______
NARA [144]

Answer:

I believe its C

6 0
4 years ago
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