3. What do platelets do?

For each set of reactions, determine the value of δh2 in terms of δh1. a+b→2c,δh1 1/2a+1/2b→c,δh2=? express your answer in terms

allochka39001 [22]

ΔH2 = - δH1 δH2 = - 2 x δH1 δH2 = 2 x <span>δ</span>H1

3 0

3 years ago

Gloria is making a model of an atom. She uses three different colors to represent the three basic particles that make up the ato

xz_007 [3.2K]

Answer:

protons and neutrons

Explanation:

The nucleus of the atom contains protons and neutrons. The electrons of the atom orbit the nucleus.

8 0

3 years ago

An atom's Lewis dot structure has four dots. Which of the following elements could it be, and why?

Elis [28]

Answer:

Beryllium, because it is in period 2 and has four total electrons.

Explanation:

7 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

Read 2 more answers

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl

Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

$\Delta S = S(products)-S(reactants)----(1)$

a) $PCl3(l) + Cl2(g) \rightarrow PCl5(s)$

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) $2HgO(s) \rightarrow 2Hg(l) + O2(g)$

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) $H2(g) \rightarrow 2H(g)$

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) $U(s) + 3F2(g) \rightarrow UF6(s)$

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

8 0

3 years ago