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kaheart [24]
3 years ago
10

Research and describe one career in health and fitness that you would consider

Chemistry
2 answers:
kiruha [24]3 years ago
6 0
I would research the job of being a fitness coach, that is very popular in health and fitness
Sati [7]3 years ago
3 0

Answer:

a PE teacher would be in both catagories  of heath and fitness

Explanation:

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How many molecules are in 5 moles of O2?
SVETLANKA909090 [29]

Answer: 6.02 × 10^24

Explanation:

7 0
3 years ago
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PLEASE HELP WITH THIS QUESITION WILL GIVE BRAINLIST 2 BEST ANSWER.
Lera25 [3.4K]

Answer: show all

Explanation:

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4 0
3 years ago
a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is
fiasKO [112]

Answer: The empirical formula is CH_2.

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.142}{0.142}=1

For H =\frac{0.287}{0.142}=2

The ratio of C: H = 1: 2

Hence the empirical formula is CH_2.

4 0
3 years ago
When energy in the atom is released, what occurs?
sineoko [7]

Answer:

What happens when electrons in atoms absorb or release energy? When electrons absorb or release energy, their electrons can move to higher or lower energy levels. These electrons lose energy by emitting light when they return to lower energy levels.

Explanation:

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4 0
3 years ago
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
3 years ago
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