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Ann [662]
3 years ago
5

Dessert biomes have all of these EXCEPT-

Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0
Answer: B). little wind


Explanation: Deserts have few large animals, fertile soil, and little rainfall but they have lots of wind
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What is the ground state electron configuration for Ru2+?
Natali5045456 [20]

Electronic configuration: The distribution or arrangement of electrons of a molecule or an atom in molecular or atomic orbitals.

Ground state electron configuration: The distribution of electrons of an atom or molecule around the nucleus with lower levels of energy.

Now, Ru stands for Ruthenium with atomic number 44. It is a metal and thus, has ability to lose electrons and, becomes positively charged ion.

One can write the electronic configuration with the help of atomic number and Afbau principle, Pauli exclusion principle etc.

Ground electronic Configuration is as follows:

Ru = s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^{7}

Soft Hand notation: [Kr]4d^{7}5s^{1}

Now, when ruthenium loses two electrons then it becomes Ru^{+2}, thus electron configuration becomes

Soft Hand notation: [Kr]4d^{6}

The ground state electronic configuration of Ruthenium is [Kr]4d^{7}5s^{1} and when it loses two electrons, then electronic configuration becomes [Kr]4d^{6} ( Ru^{+2})





5 0
3 years ago
Read 2 more answers
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



5 0
2 years ago
Read 2 more answers
How many moles of Mg are contained in 11.0<br> grams of magnesium?
DanielleElmas [232]

Answer:

Should be 0.6106 though i could be wrong

Explanation:

3 0
3 years ago
Lists how many valence electrons are shared, gained, or lost by each atom of the 6 compounds?      
Blababa [14]
<span>The number of electrons in an atom's outermost valence shell governs its bonding behavior.

In N</span>₂, three electrons are being shared by each nitrogen atom, making a total of 6 shared electrons.

In CCl₄, 4 electrons are being shared by each carbon atom and 1 electron is being shared by each chlorine atom

In SiO₂, 4 electrons are being shared by each silicon atom and 2 electrons are being shared by each oxygen atom.

In AlCl₃, 3 electrons are being shared by each aluminum atom and 1 electron is being shared by each Cl atom

In CaCl₂, 2 electrons are lost by the calcium atom and 1 electron is gained by each chlorine atom

In LiBr, 1 electron is lost by the lithium atom and 1 electron is gained by the bromine atom
6 0
3 years ago
A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther
sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0
3 years ago
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