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3 years ago

What experimental evidence did Thomson have for each statement?

2 answers:

6 0

Answer : In the experiment of Cathode Ray, Scientist J.J. Thomson found that a stream of electricity was attracted to the positive portion of an electrical plate, which helped him to conclude that there must be a negative aspect to matter. He called this discovery as the electron.

He was the one who suggested that these small, negatively charged particles were actually subatomic particles. Which are now called as "electrons".

Since atoms are found to be neutral, when it contain negatively charged electrons it must also contain some positively charged material.

Minchanka [31]3 years ago

5 0

<span>Thomson studied electric discharge in a vacuum and found that the deflection of rays was evidence of atoms containing much smaller particles. He calculated that these particles would have a large charge in relation to their mass. While he did not name electrons, he knew they existed.</span>

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kipiarov [429]

The answer is ATOMS, because matter is a solid, liquid, or gas…

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3 years ago

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Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole

trapecia [35]

Answer : The enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol$

Therefore, the enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

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4 years ago

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galina1969 [7]

Answer:

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Explanation:

H2O

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3 years ago

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Misha Larkins [42]

Answer:

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3 years ago

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

Equilibrium concentration of $NO_2$ = $\frac{22.5g}{46g/mol\times 9.5L}=0.051M$

Equilibrium concentration of $H_2O$ = $\frac{189.0g}{18g/mol\times 9.5L}=1.10M$

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 $K_c$

For the given chemical reaction:

$2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)$

The expression for $K_c$ is written as:

$K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}$

$K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}$

$K_c=3.72$

Thus the value of the equilibrium constant Kc for this reaction is 3.72

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3 years ago