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4 years ago

Which particle decomposes to produce beta radiation?

Chemistry

2 answers:

TiliK225 [7]4 years ago

8 0

Answer:

Neutron

Explanation:

guajiro [1.7K]4 years ago

6 0

The particle that decomposes to produce beta radiation is NEUTRON.

The neutron decays (udd) into a proton (uud), an electron and an anti-neutrino. This process is known as neutron beta decay and here is how it happens.

The neutron that has a charge of zero is made up of up and two down quarks. One of the down quarks is transformed into an up quark. Since the down quark has a -1/3 charge and the up quarks has a positive charge of 2/3, it immediately follows that this process is mediated by a virtual W particle which carries away a negative charge. The new up quark rebounds away from the emitted W-. Then comes the formation of a proton. An electron and antineutrino them emerge from the virtual W-boson. The proton, electron, and antineutrino will finally move away from one another,

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In a __________________ reaction, elements are not changed, just rearranged. Small amounts of energy take place.

emmainna [20.7K]

D. In a Chemical reaction, elements are not changed, just rearranged. Small amounts of energy take place.

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3 years ago

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An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI prese

Maru [420]

Answer: The molarity of $Na_2S_2O_3$ is 0.108 M

Explanation:

$KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2$

$2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

$\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol$

1 mole of $KIO_3$ produces = 3 moles of $I_2$

$2.8\times 10^{-4}$ moles of $KIO_3$ produces = $\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}$ moles of $I_2$

Now 1 mole of $I_2$ uses = 2 moles of $Na_2S_2O_3$

$8.4\times 10^{-4}$ moles of $I_2$ uses = $\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}$ moles of $Na_2S_2O_3$

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M$

The molarity of $Na_2S_2O_3$ is 0.108 M

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3 years ago

What did timbuktu contributed to mali's important as a kingdom

Alex73 [517]

Timbuktu was considered a MASSIVE trade center in Mali. That is how all of the foreign goods came into the kingdom. It also had education centers which was very important in Mali.

5 0

4 years ago

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[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)

pav-90 [236]

Answer:

For A: The expression for $K_{eq}$ is given below.

For B: The value of $K_{eq}$ at 25°C is 0.0185512

For C: The value of $K_{eq}$ at 65°C is 0.2887886

For D: The reaction is endothermic in nature.

Explanation:

For A:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For the given chemical reaction:

$[Co(H_2O)_6]^{2+}(aq.)+4Cl^-(aq.)\rightleftharpoons [CoCl_4]^{2-}(aq.)+6H_2O(l)$

The expression of $K_{eq}$ for above equation without the concentration of liquid water is:

$K_{eq}=\frac{[CoCl_4]^{2-}}{[Co(H_2O)_6]^{2+}[Cl^-]^4}$ ......(1)

The expression is written above.

For B:

We are given:

$[CoCl_4]^{2-}=0.0334612M$

$[Co(H_2O)_6]^{2+}=0.966539M$

$[Cl^-]=1.86616M$

Putting values in equation 1, we get:

$K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512$

Hence, the value of $K_{eq}$ at 25°C is 0.0185512

For C:

We are given:

$[CoCl_4]^{2-}=0.234625M$

$[Co(H_2O)_6]^{2+}=0.765375M$

$[Cl^-]=1.06150M$

Putting values in equation 1, we get:

$K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886$

Hence, the value of $K_{eq}$ at 65°C is 0.2887886

For D:

For Endothermic reactions, $\Delta H>0$ , which is positive

For Exothermic reactions, $\Delta H$ , which is negative

To calculate $\Delta H$ of the reaction, we use Van't Hoff's equation, which is:

$\ln(\frac{K_{65^oC}}{K_{25^oC}})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{65^oC}$ = equilibrium constant at 65°C = 0.2887886

$K_{25^oC}$ = equilibrium constant at 25°C = 0.0185512

$\Delta H$ = Enthalpy change of the reaction = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $25^oC=[25+2730]K=298K$

$T_2$ = final temperature = $65^oC=[65+2730]K=338K$

Putting values in above equation, we get:

$\ln(\frac{0.2887886}{0.0185512})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{298}-\frac{1}{338}]\\\\\Delta H=57471.26J/mol$

As, the calculated value of $\Delta H>0$ . Thus, the reaction is endothermic in nature.

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3 years ago

Are bath bombs pure substances or mixtures

mezya [45]

Answer:

Yes.

Explanation:

Bath bombs are usually made from three key ingredients: baking soda, citric acid and cornstarch, said Frankie Wood-Black, an instructor in chemistry at Northern Oklahoma College and an experienced bath-bomb maker. Often, bath bombs also include dyes and perfumes, and sometimes they have epsom salt.

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3 years ago

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