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Alex_Xolod
2 months ago
14

What are the respective central-metal oxidation state, coordination number, and overall charge on the complex ion in Na2[Cr(NH3)

2 (NCS)4]?
Chemistry
1 answer:
lions [2.9K]2 months ago
6 0

Answer:

Central metal oxidation state: +2

Coordination number: 6

Overall charge: -2

Explanation:

For the ion complex:

Na₂[Cr(NH₃)₂(NCS)₄]

The central metal is chromium, with NH₃ and NCS as the ligands.

NH₃ acts as a neutral ligand, while NCS carries a negative charge.

The entire complex has a charge of:

2Na⁺ +  [Cr(NH₃)₂(NCS)₄]⁻² → -2

Since each NCS contributes -1 and there are four NCS, the Cr must possess an oxidation state of +2 to achieve an overall charge of -2.

With 2 NH₃ and 4 NCS attached, the coordination number sums to 2+4 = 6

I trust this clarifies the matter!

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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
eduard [2782]

Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

E = h \nu

1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

\nu = 0.163 \times 10^{17} s^{-1}

or, \nu = 1.63 \times 10^{16} s^{-1}

It is known that \nu = \frac{c}{\lambda}

1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}

\lambda = 1.84 \times 10^{-8} m

According to the De-Broglie equation \lambda = \frac{h}{p}

with p = m \times \nu

So, \lambda = \frac{h}{m \times \nu}

m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m} = 3.6 \times 10^{-26} J/m

Squaring both sides gives us:

(m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}

12.96 \times 10^{-52} = m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

where m = mass of the electron

Therefore, m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

=\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

=1.42 \times 10^{-21} J

Since K.E = \frac{1}{2}m \nu^{2}

= \frac{1.42 \times 10^{-21} J}{2}

=0.71 \times 10^{-21} J

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is 7.1 \times 10^{-22} J.

4 0
2 months ago
doctor has ordered that a patient be given 20 g of glucose, which is available in a concentration of 70.00 g glucose/1000.0 mL o
lions [2927]
The patient needs to receive 285.71 ml.
1000 ml contains 70 gr of glucose.
   x          contains 20 gr of glucose.

x=1000*20/70
8 0
1 month ago
Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
eduard [2782]

Answer:

The integer value of x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molar concentration of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Weight of hydrated sodium carbonate = n = 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

By solving for x, we arrive at:

x = 9.95, approximating to 10

The integer x in the hydrate equals 10.

6 0
2 months ago
How many molecules of PF5 are found in 39.5 grams of PF5?
Anarel [2989]

Response:

1.9 \times 10^{23} molecules of PF_5 can be found in 39.5 grams of PF_5.

Clarification:

Atomic weights: P= 31, F= 19,

The molar mass equals 1 atomic weight of P + 5 atomic weights of

 F= 31+5 × 19\times = 31+95

=126 g/mole

The number of moles in 39.5 gm of

equals \frac{Mass}{Molar mass}

 = \frac{39.5}{126}moles

 =0.3134 moles

1 mole of any substance encompasses

0.3131 moles comprises 0.3134

= 1.9 \times 10^{23} molecules

Thus, 1.9 \times 10^{23} molecules of PF_5 can be found in 39.5 grams of PF_5.

7 0
1 month ago
Why would it be better to be an r-selected species if the water resources in an area were to become more limited over a short pe
lions [2927]
An r-selected species has a significantly faster reproductive rate compared to K-selected species.

The focus of r-selected species is on quick maturation and reproduction. They are likely to breed during short periods when water supply is available, thus enhancing their survival chances.

Conversely, K-selected species prioritize nurturing their young and tend to reproduce later. Due to the longer maturation time before breeding, by the time K-selected species are ready, the water supply may be depleted, leading to lower survival odds.

Hope this clarifies!
If you have more questions, don’t hesitate to ask! Have a fantastic day!:)

~Collinjun0827, Junior Moderator
6 0
2 months ago
Read 2 more answers
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