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stiks02 [169]
3 years ago
14

The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?

Physics
1 answer:
Likurg_2 [28]3 years ago
4 0
\frac{7.85g }{1.59 \frac{g}{ cm^{3} } } = 4.937 cm^{3}
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An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value
Dimas [21]

Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

R = Rydberg constant, R=1.097\times 10^7\ m^{-1}

\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

Solving above equation we get the value of final n is,

n_f=2.04

or

n_f=2

So, it will relax in the n = 2. Hence, this is the required solution.        

6 0
3 years ago
Read 2 more answers
I push a box with mass 10 kg across a 15 m room. If I apply a force of 20N to do this how much work is being done on the box?
devlian [24]
You could use the formula

W=Fd

F(force)=20N

D(distance/displacement) =15m

W=(20N)(15m)

W= 300 J
8 0
3 years ago
Who was this scientist, what ideas did he form, and how did he figure out these new ideas of atoms?
saul85 [17]

Answer:

Atoms cannot be divided.

Explanation:

5 0
3 years ago
The student places 0.5 kg of potato into a pan of water.
nasty-shy [4]

Answer:

136000 J or 136 kJ

Explanation:

Formula

Heat = m * c * deltaH

Givens

m= 0.5 kg

c = 3400 J / (kg * oC)

Deltat = (100oC - 20oC)

deltat = 80oC

Solution

Heat = 0.5 kg * 3400 J/(kg* oC) * 80oC

Heat = 136000 Joules

Heat = 136 kg

Technically there is only 1 place of accuracy.

6 0
2 years ago
Physical science 9th grade peeps helpppo
kakasveta [241]
It's the 3rd one bc they are produced by beta decay
3 0
2 years ago
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