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3 years ago

9

Elements are distinguished from each other by the number of protons present in their nuclei??

1 answer:

Kisachek [45]3 years ago

7 0

False....................

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If 3 moles of a compound use 24 J of energy in a reaction, what is the A<br> Hreaction in J/mol?

erma4kov [3.2K]

Answer:

+8 J/mol

Explanation:

The following were obtained from the question:

Mole of the compound = 3 moles

Heat Energy used = 24J

Hreaction =?

The heat of reaction is the energy released or absorb during a chemical reaction. It is can obtained mathematically by the following equation:

Hreaction = Heat energy/mole

With the above equation, we can easily find the Hreaction for the question given as follow:

Hreaction = Heat energy/mole

Hreaction = 24J / 3mol

Hreaction = 8 J/mol

Since the compound used the energy, the Hreaction is +8 J/mol

8 0

3 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

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2 years ago

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I might be small, but I manufacture most of the protein the cells needs

solniwko [45]

I’m soooo confusing with what your question is lol

7 0

3 years ago

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Describe the shapes and relative energies of the s,p,d, and f atomic orbitals?

Mariulka [41]

Relative energies are in the order:

s < p < d < f

And the shape of these orbitals are

s - spherical

p - dumbbell

d - double dumbbell

f - double double dumbbell

6 0

3 years ago

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Balance the equation AlCl3 + H2SO4 yields Al2(SO4)3 +HCl

Natalija [7]

Answer:

2AlCl3 + 3H2SO4 → Al2(SO4)3 + 6HCl

Explanation:

3 0

3 years ago