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3 years ago

Nuclei may be unstable if they have

1 answer:

3 0

Nuclei may be unstable if they have
a too much energy.
b too many protons.
C an unstable ratio of protons to neutrons.
d any of the above.

The answer would be, D any of the above.
Hope this helps

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Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magne

Fynjy0 [20]

Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.

(a) Barium nitrate is a salt formed by the cation barium Ba²⁺ and the anion nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a salt formed by the cation potassium K⁺ and the anion chlorate ClO₃⁻. Its formula is KClO₃.

(b) The balanced equation for the decomposition of potassium chloride is:

2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)

Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)

(d) The balanced equations of metals with oxygen to form metal oxides are:

2 Mg(s) + O₂(g) ⇄ 2 MgO(s)

4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)

4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)

5 0

3 years ago

Distinguish between deliquescence and efflorescence.

netineya [11]

Explanation:

Deliquescent substances are solids that absorb moisture from the atmosphere until they dissolve in the absorbed water and form solutions. Efflorescent: Efflorescent substances are solids that can undergo spontaneous loss of water from hydrated salts.

5 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

3 years ago

What 2 instruments are commonly used to measure air pressure?

levacccp [35]

I think it is the barometer and aneroid barometer

8 0

3 years ago

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

slega [8]

Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted. = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$Binding\ energy=43.43\times 10^{-20}\ J$

5 0

3 years ago