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madreJ [45]
3 years ago
12

What characteristic do all parts of the electromagnetic spectrum have in common

Chemistry
1 answer:
Elza [17]3 years ago
5 0
All electromagnetic waves are amplitude, a characteristic frequency and wavelength, and the ability to travel through a vacuum at the same speed (the speed of light)
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Can you identify 2 characteristics of combustion reactions?
monitta
An element or compound will react with oxygen and will produce carbon dioxide, water, and sometimes carbon (if it is an incomplete combustion).
4 0
3 years ago
Leah is trying to determine how music affects the growth of plants. She hypothesizes that classical music will cause a plant to
Pani-rosa [81]
Leah's experiment:
A. Takes careful and regular measurements (she measures the growth every day)
B. Has a specific hypothesis and a controlled experiment (the music is the only changing variable, the growth conditions are identical, and this is what she wants to test in her hypothesis)
C. Experimental data that can be recorded and analyzed (measuring the growth of plant fits this)

So the answer must be D, since there is no mention of past research that scientists have found on this topic.
8 0
3 years ago
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5. An element whose highest occupied s sublevel and a nearby d sublevel <br> contain electrons
Mumz [18]

Answer:

                    Transition Element

Explanation:

                     Transition elements are defined as those elements which can form at least one stable ion and has partially filled d-orbitals. They are also characterized by forming complex compounds and having different oxidation states for a single metal element.

                      Transition metals are present between the metals and the non metals in the periodic table occupying groups from 3 to 12. There general electronic configuration is as follow,

                                                   (n-1)d ¹⁻¹⁰ns ¹⁻²

The general configuration shows that for a given metal, the d sublevel will be in lower energy level as compared to corresponding s sublevel. For example,

Scandium is present in fourth period hence, its s sublevel is present in 4rth energy level so its d sublevel will be present in 3rd energy level respectively.

Hence, we can conclude that for transition metals the electron are present in highest occupied s sublevel and a nearby d sublevel .

5 0
3 years ago
Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K
yawa3891 [41]

Answer:

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

Explanation:

Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)

Concentration of nickel ion = [Ni^{2+}]=x

Concentration of nickel complex= [[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L

Concentration of ethylenediamine = [en]=\frac{0.80 mol}{2 L}=0.40 mol/L

The formation constant of the complex = K_f=4.0\times 10^{18}

The expression of formation constant is given as:

K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}

4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}

x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}

x=3.125\times 10^{-19} mol/L

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

6 0
3 years ago
A certain radioactive isotope decays at a rate of 0.2​% annually. Determine the ​half-life of this​ isotope, to the nearest year
pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau}

Where:

m - Current isotope mass, measured in kilograms.

t - Time, measured in years.

\tau - Time constant, measured in years.

The solution of this differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%

1 - \frac{m(t+1)}{m(t)} = 0.002

1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002

1 - e^{-\frac{1}{\tau} } = 0.002

e^{-\frac{1}{\tau} } = 0.998

-\frac{1}{\tau} = \ln 0.998

The time constant associated to the decay is:

\tau = -\frac{1}{\ln 0.998}

\tau \approx 499.500\,years

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

t_{1/2} = \tau \cdot \ln 2

If \tau \approx 499.500\,years, the half-life of the isotope is:

t_{1/2} = (499.500\,years)\cdot \ln 2

t_{1/2}\approx 346.227\,years

The half-life of the radioactive isotope is 346 years.

6 0
2 years ago
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