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3 years ago

What change occurs to an atom when it forms negative ion

Chemistry

2 answers:

Artist 52 [7]3 years ago

6 0

Answer:

The outer shells of non-metal atoms gain electrons when they form ions. The ions formed are negative because they have more electrons than protons

Explanation:

IrinaVladis [17]3 years ago

6 0

Ions are formed when atoms gain or lose electrons. Gaining electrons changes an atom into a negative ion. Losing electrons changes an atom into a positive ion. Individual atoms often do not form ions by themselves.

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Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you

oee [108]

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

$\eta_{KClO_{3}} = \frac{m}{M}$

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

$\eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles$

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

$\eta_{O_{2}} = \frac{3}{2}*0.252 moles = 0.378 moles$

Finally, the mass of O₂ is:

$m = 0.378 moles*32 g/mol = 12.10 g$

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

6 0

2 years ago

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

6 0

3 years ago

A gas fills a balloon at a temperature of 27 C and 101.325 kPa of pressure. What will the pressure of the balloon be if the gas

JulijaS [17]

135.1‬kPa

Explanation:

Given parameters:

T1 = 27°C

P1 = 101.325 kPa

T2 = 127°C

Unknown:

P2 = ?

Solution:

Using a derivative of the combined gas law where we assume that the gas has a constant volume, we can solve for the unknown.

At constant volume:

$\frac{P1}{T1} = \frac{P2}{T2}$

P1 is the initial pressure

T1 is the initial temperature

P2 is the final pressure

T2 is the final temperature

Take the given temperature to K

T1 = 27 + 273 = 300K

T2 = 127 + 273 = 400K

Input the variables:

$\frac{101.325}{300} = \frac{P2}{400}$

P2 = 135.1‬kPa

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

3 0

3 years ago

If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give

tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

The maximum height attained by nitrogen gas molecule = h
The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

$K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J$

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

The potential energy at height h = $P.E = 6.381\times 10^{-21} J$
Molar mass of nitrogen gas = 28.0134 g/mol
Mass of nitrogen gas molecule = m

$m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg$

The acceleration due to gravity = g = 9.8 m/s^2
The maximum height attained by nitrogen gas molecule = h
The potential energy is given by:

$P.E=mgh$

$6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km$