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3 years ago

15

A chemical reaction occurs between Calcium, Oxygen, and Chlorine. The reaction is demonstrated:

2 answers:

ad-work [718]3 years ago

8 0

Answer:

D.

Double replacement, CaO + Cl2O

Explanation:

DaniilM [7]3 years ago

8 0

Answer:

Double Replacement, Ca0 + CI20

Explanation:

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volume of the displaced water = 600g/19.3g/cm3
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The discharge of phosphate in detergents to the environment has led to imbalances in the life cycle of freshwater lakes. A chemi

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2 years ago

Which ion would represent the ion of an element from Group 2A?

anzhelika [568]

Answer:

option X^2- as member of group 2A are non metal

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3 years ago

Suppose you have a 2.0 molar solution of sodium hydroxide (naoh), and you need 65 ml of 0.6 molar naoh. how should you make this

nalin [4]

The given molarity of sodium hydroxide solution = 2.0 M

The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH

Converting 65 mL to L:

$65mL*\frac{1L}{1000mL} =0.065L$

Calculating the moles of NaOH in the final solution:

$0.065L * \frac{0.6 mol}{L} =0.039mol$

Finding out the volume of 2.0 M solution taken to prepare the final solution:

$0.039 mol * \frac{1L}{2.0mol}=0.0195L*\frac{1000mL}{1L} =19.5mL$

Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.

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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>

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3 years ago