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Novay_Z [31]
4 years ago
13

The group in an experiment that is not exposed to the teased variable is called the _ group

Chemistry
1 answer:
wlad13 [49]4 years ago
8 0
If the conditions of the experiment are NOT changing for this group, then it's the control group. 
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AN atoms nucleus contains it's protons and electrons true or false
lesantik [10]
True, an atoms nucleus has both protons and electrons.
4 0
3 years ago
Which one has more mass a bag of cotton balls or a bag of nails and explain
Mama L [17]
You would think that the bag of nails would have more mass but their masses are identical. <span>If you were to put them both in a vacuum chamber and let them fall from a great height, they would fall the same speed. The vacuum chamber would suck all of the air out of the cotton balls, thus making it heavier and weigh the same as the bag of nails.

Hopefully this is helpful and makes sense.</span>
6 0
3 years ago
How are covalent compounds similar to acids?
Ivan

Answer:

Acids are substances that produce an create high amounts of H+ ions when dissolved in water. And because the hydrogen bond's with non-metals,it forms covalent bonds. So,all acids are covalent bonds.

Explanation:

Colavent compounds are colavent bonds

Source:

https://www.quora.com/Are-acids-covalent-compound

5 0
3 years ago
calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?
inessss [21]

<u>Answer:</u>

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}

\Delta E=-1312[0.111-1]KJ mol^{-1}

\Delta E=1166 KJ mol^{-1}

\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}

=193623 \times 10^{-23}  \frac {J}{photon}

\Delta E=1.93623 \times 10^{-18}  \frac {J}{photon}

\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}

h is planck's constant  

c is the speed of light

λ is the wavelength of light  

\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18}  J/photon)}

Wavelength

\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m}  =103 nm (Answer)

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0
3 years ago
What is the part of an experiment that is not being tested and is used for comparison
STALIN [3.7K]
The controlled variable, I think.
7 0
3 years ago
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