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3 years ago

Characteristics of α, β, and γ rays

Chemistry

1 answer:

Allisa [31]3 years ago

7 0

Answer:

1. it is symbolized as 4/2 He : Alpha rays

2. it has the weakest penetrating power : Alpha rays

3. It is a high-speed electron : beta rays

4. It possesses neither mass nor charge : gamma rays

5. it has the strongest penetrating power : beta rays

6. its is symbolized as 0/-1e : beta rays

7. it is the most massive of all the components: alpha rays

Explanation:

Let us consider the characteristics of each of the given rays.

α rays: These are helium nucleus so are symbolized by $He^{4}_{2}$

Due to two protons and two neutrons unlike beta and gamma rays these are the most massive and thus have least penetrating power among the three given rays.

β rays: These are actually high speed electrons and are symbolized as $^{0}_{-1}e$ . Due to lesser mass than alpha rays they are more penetrating than them however less penetrating than gamma rays.

γ rays : They carry no charge or mass. Due to least massive among the three rays they have highest penetrating power.

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Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

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3 years ago

100 points right here first come first serve. amos thejonah2016

Kisachek [45]

Answer:

siya at ako ayieeeeeeeeee

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3 years ago

what happens when a gas obtained by heating Ammonium Chloride and slaked lime are passed through copper sulphate ?

gogolik [260]

Answer:

Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.

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3 years ago

List and explain the four most important factors that control the relative reactivity of a carbonyl-containing functional group

ioda

Resonance, leaving group, carbonyl carbon delta+, and steric effect is the most crucial variables that affect the relative reactivity of a functional group containing a carbonyl in an addition or substitution process.

Discussion:

1. Carbonyl Carbon Delta+: The carbonyl group becomes more electrophilic and accelerates nucleophilic assault when the carbonyl carbon delta+ is bigger.

2. Resonance: When the carbonyl is transformed into the tetrahedral adduct, it may be lost. Loss of resonance increases the energy of the transition state for this nucleophilic assault because resonance has the function of stabilizing. Therefore, a carbonyl functional group's resistance to nucleophilic attack increases as resonance in the group increases in importance.

3. Leaving group: Tetrahedral adduct fragmentation is encouraged by a better LG.

4. Steric effects: The nucleophilic attack on carbonyl carbon is delayed when sterically impeded.

Learn more about carbonyl here:

brainly.com/question/21440134

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1 year ago

A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment,

nikitadnepr [17]

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

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2 years ago