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posledela
2 years ago
9

If radioactive caesium was reacted with chlorine, would you expect the caesium chloride produced to be radioactive? Explain you

answer.
Chemistry
2 answers:
CaHeK987 [17]2 years ago
8 0

Answer:  The answer is YES.

Explanation:  Isotopic substitution is the kind of substitution in which the present active species is replaced by its isotope in order to check the reactivity of the given species.

Radioactive substitution is also the same kind of substitution in which the active species is replaced by its radioactive isotope.

Thus, in cesium chloride when radioactive cesium is reacted with the chlorine, then the product formed cesium chloride is expected to be radioactive because radioactivity or isotopic substitution has nothing to do with the bonding. Only outermost electrons of the cesium has been transferred from cesium to chlorine , and its nucleus has not been affected in the same process.

aniked [119]2 years ago
3 0
Yes due to the radioactivity having nothing to do with the chemical equation given it will release radiation at a rate determined by it's half life.
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What is the formal charge on the nitrogen in hydroxylamine, h2noh?
andrew11 [14]
<h3>Answer:</h3>

             Formal Charge on Nitrogen is "Zero".

<h3>Explanation:</h3>

Formal Charge on an atom in molecules is calculated using following formula;

Formal Charge  =  [# of Valence e⁻s] - [e⁻s in lone pairs + 1/2 # of Bonding e⁻s]

As shown in attached picture of Hydroxylamine, Nitrogen atom is containing two electrons in one lone pair of electrons and six electrons in three single bonds with two hydrogen and one oxygen atom respectively.

Hence,

                                  Formal Charge  =  [5] - [2 + 6/2]

                                  Formal Charge  =  [5] - [2 + 3]

                                  Formal Charge  =  5 - 5

                                  Formal Charge  =  0    (zero)

Hence, the formal charge on nitrogen atom in hydroxylamine is zero.

5 0
1 year ago
Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a
Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

3 0
2 years ago
Both black and white road surfaces radiate energy. at midnight on a starry night the warmer road surface is the
luda_lava [24]
The black road because during the day it absorbed more radiation than the with one
5 0
1 year ago
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Paul [167]

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

7 0
2 years ago
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
2 years ago
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