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Alex_Xolod [135]
2 years ago
9

An iron loses electrons. State the type of charge it will end up with

Chemistry
1 answer:
erastova [34]2 years ago
4 0

Answer:

Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic

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Given the thermochemical equations below, What is the standard heat of formation of CuO(s)? 2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆
OlgaM077 [116]

-130KJ is the standard heat of formation of CuO.

Explanation:

The standard heat of formation or enthalpy change can be calculated by using the formula:

standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation

Data given:

Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ

2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ

CuO + Cu ⇒ Cu2O (-11.3 KJ)      ( Formation of Cu2O)

When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.

Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ

So standard heat of formation of  formation of Cu0 as:

Cu + 1/2 02 ⇒ CuO

putting the values in the equation

ΔHf = ΔH1 + ΔH2     (ΔH1 + ΔH2  enthalapy of reactants)

heat of formation = -11.3 + (-119.35)

                            = - 130.65kJ

-130.65 KJ is the heat of formation of CuO in the given reaction.

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Question 7 of 32
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Answer:O D to protect the environment from harmful chemicals

Explanation:

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Exposure to what type of radiant energy is sensed by human skin as warmth?
larisa86 [58]
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Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

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the density of a black hole is 22.61 g/cm3 i think

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