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2 years ago

15

How many phosphorus atoms are there in 1.75 mol of calcium phosphate, Ca3(PO4)2

1 answer:

elena-s [515]2 years ago

6 0

Answer:

It has 2 phosphorus atoms in 1 mol

so for 1.75 mol =2×1.75

Explanation:

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How many electrons in an atom can share the quantum numbers n = 4 and l = 3?

o-na [289]

Answer:

$\boxed{\text{14}}$

Explanation:

If l = 3, the electrons are in an f subshell.

The number of orbitals with a quantum number l is 2l + 1, so there

are 2×3 + 1 = 7 f orbitals.

Each orbital can hold two electrons, so the f subshell can hold 14 electrons.

$\boxed{\textbf{14 electrons}} \text{ can share the quantum numbers n = 4 and l = 3.}$

3 0

3 years ago

Estimate the standard internal energy of formation of liquid methyl acetate (methyl ethanoate, ch3cooch3) at 298 k from its stan

LenKa [72]

solution:

$the reaction for formation of methyl acetate is\\CH_{3}OH+CH_{3}COOH----------->CH_{3}COOCH_{3}+H_{2}O\\standard internal energy \\\delta u=\delta H -\delta(pv)=\delta H -\delta n\times RT\\where \delta n=change in number of moles=0\\\delta u=\delta H\\=-442kj/mol$

6 0

2 years ago

Read 2 more answers

Calculate the mass of one mole of these substances.

pogonyaev

Answer:

a. 53.5 g/mol

b. 80.06 g/mol

c. 133.33 g/mol

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

Reading a Periodic Table
Molar Mass - 1 mol per <em>x</em> grams substance

Explanation:

<u>Step 1: Define</u>

a. NH₄Cl

b. NH₄NO₃

c. AlCl₃

<u>Step 2: Find masses</u>

Molar Mass of N - 14.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of Cl - 35.45 g/mol

<u>Step 3: Calculate compound masses</u>

Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol

Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol

Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol

6 0

3 years ago

Calculate the standard free-energy change at 25 ∘C for the following reaction:

lianna [129]

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$ ( $\Delta G^{0}$ ): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

8 0

2 years ago

Read 2 more answers

Jose mixed two clear liquids together in a beaker in equal amounts. Jose measured the mass. The mass of the two combined solutio

Dmitrij [34]

Who cares. Joe can solve his own problem instead of making people do it. (This was not toward you. This was supposed to be funny)

5 0

2 years ago