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Solnce55 [7]
2 years ago
11

For each of the following ground state electron configurations, determine what is incorrect. N: 1s²2s¹2p³ A) The configuration i

s not in the ground state. B) The configuration is missing an electron in the 2s orbital. C) The configuration has too many electrons in the 2p orbital.
Chemistry
1 answer:
MatroZZZ [7]2 years ago
6 0

Answer:

B

Explanation:

It should be 2s² which means the answer is the configuration is missing an electron in the 2s orbital.

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What type of reaction is the digestion of solid copper wire by nitric acid?
Nuetrik [128]

Copper nitrate and nitric oxide are produced in this reaction.

5 0
2 years ago
Which statement best describes how electrons fill orbitals in the periodic table? Electrons fill orbitals in order of their incr
Stolb23 [73]
Electrons fill orbitals in order of increasing energy from left to right. As the group number increase also the number of valence electorns of each group will increases
5 0
2 years ago
Read 2 more answers
During a lab experiment performed at STP conditions, you prepare HCl by reacting 100. ml of Cl2 gas with an excess of H2 gas.
Brrunno [24]

Answer: 19.4 mL Ba(OH)2

Explanation:

H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)

At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.

0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2

Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.

0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl

HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.

0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2

3 0
2 years ago
Write a balanced equation for the transmutation that occurs when a scandium-48 nucleus undergoes beta decay.
Tju [1.3M]

Answer:

A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.

\rm ^{48}_{21}Sc \to ^{48}_{22}Ti + ^{\phantom{1}\,0}_{-1}e^{-} + \bar{\mathnormal{v}}_e.

Explanation:

There are two types of beta decay modes: beta-minus and beta-plus.

In both decay modes, the mass number of the nucleus stays the same.

However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.

Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.

Look up the atomic number and relative atomic mass for the element scandium.

  • The atomic number of \rm Sc is 21.
  • The relative atomic mass of \rm Sc is approximately 45.0.

This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.

Each element (e.g, \rm Sc) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average  across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.

The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.

The atomic number of the nucleus will increase by 1. 21 + 1 = 22. That corresponds to titanium. The mass number stays the same at 48. Hence the daughter nucleus would be titanium-48. Note that two other particles: one electron and one electron \rm e^{-} and one antineutrino \bar{v}_{\text{e}} (note the bar.) The neutrino helps balance the lepton number of this reaction.

6 0
2 years ago
Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:
Katarina [22]

Answer and Explanation:

The 3 reactions represented are

C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O

C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O

C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O

ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol

ΔH°f = ΔH°(products) - ΔH°(reactants)

For reaction 1,

ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol

For reaction 2,

ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol

For reaction 3,

ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol

b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.

Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol

Number of moles = 254/296 = 0.858 mole

heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol

c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole

5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole

The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole

q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ

QED!!!

8 0
2 years ago
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