Purchasing insurance can help Adrian risk. Adrian’s best decision in this case is to because the policy is

alexdok [17]

The name of the compound K3N is potassium nitride (C).

8 0

3 years ago

If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a

Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0

3 years ago

Who holds the public property rights

Citrus2011 [14]

The answer to your question is,

B. Government.

-Mabel <3

3 0

3 years ago

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Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)

Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(c) K⁺ (charge 1+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0

3 years ago

In the animation, you can see that the electrons occupy different orbitals according to the energy level of each orbital. A sing

lara31 [8.8K]

B) The He atom has two electrons that have parallel spin in its 1s orbital.

C) Electrons generally occupy the lowest energy orbital first.

D) The arrangement of the orbitals is the same in a multielectron atom and a single-electron atom

E) The C atom has two unpaired electrons.

Explanation:

In writing sublevel notations certain rules are followed:

The maximum number of electrons in the orbitals of sublevels are:

two for s-sublevel with one orbital

six for p-sublevel with three orbitals

ten for d-sublevel with five orbitals

fourteen for f-sublevel with seven orbitals

Aufbau's principle states that sublevels with lower energies are filled before those with higher energies.

1s 2s 2p 3s 3p 4s 3d 4p 5s increasing order of filling

Pauli exclusion principle states that no two electrons can have the same set of the four quantum numbers i.e electrons in the same orbital cannot spin in the same direction.
Hund's rule states that electrons go into degenerate orbitals singly first before pairing up.

Based on these principles and the animation, the following options are correct:

B) The He atom has two electrons that have parallel spin in its 1s orbital.

Helium has two electrons in the 1s orbital. From Pauli's exclusion principle, we see that the two electrons cannot spin in the same way. Therefore, they have parallel spin

C) Electrons generally occupy the lowest energy orbital first.

From Aufbau's principle, electrons go into orbitals with lower energies first before those with higher energies.

D) The arrangement of the orbitals is the same in a multi-electron atom and a single-electron atom

1s 2s 2p 3s 3p 4s 3d 4p 5s

Orbitals are arranged in the order of their energies from the lowest to the highest.

E) The C atom has two unpaired electrons.

Configuration for carbon atom is shown below:

1s² 2s² 2p²

Learn more:

Orbitals brainly.com/question/1832385

#learnwithBrainly

8 0

3 years ago

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