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bogdanovich [222]

4 years ago

11

How many atoms are in 1.4 mol of phosphorus trifluoride (PF3)?

1 answer:

Trava [24]4 years ago

6 0

One mol is equal to the Avogadro's Number. Using this equivalence, we can obtain the number of molecules contained in 1.4 mol:

$1.4\ mol\ PF_3\cdot \frac{6.022\cdot 10^{23}\ molec}{1\ mol} = 8.43\cdot 10^{23}\ molec\ PF_3$

Each molecule contains 4 atoms: three atoms of fluorine and one atom of phosphorus:

$8.43\cdot 10^{23}\ molec\ PF_3\cdot \frac{4\ atoms}{1\ molec\ PF_3} = \bf 3.37\cdot 10^{23}\ atoms$

Is there any mistake in my English? Please let me know!

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Is lithium (Li) more or less reactive than beryllium (Be)? Why?

balu736 [363]

Answer:

Lithium is more reactive as it has only one electron in its valance shell. Beryllium has 2 valance electrons. Lithium is more reactive as it is easier to lose oneelectron than losing two electrons.

3 0

3 years ago

Read 2 more answers

The maximum number of electrons in each type of sublevel (s, p, d, and f, respectively

Soloha48 [4]

S = 2
p = 6
d =10
f = 14

6 0

3 years ago

lutik1710 [3]

Answer:

See Explanation

Explanation:

8. 3, 1, -1, +1/2 => 3pₓ¹ => Aluminum

9. 4, 2, +1, +1/2 => 4d₁¹ => Chromium

10. 6, 1, 0, -1/2 => 6p₀² => Argon

11. 4, 3, +3, -1/2 => 4f₊₃² => Lutetium

12. 2, 1, +1, -1/2 => 2p₊₁² => Neon

8 0

3 years ago

What will be the effect on the solubility of BaF2 if the following changes are made: increasing temperature; adding NaF; adding

borishaifa [10]

In BaF₂ the solubility will decreases on adding NaF

the solubility will increases on adding HCl

<h3>SOLUBILITY OF BARIUM FLUORIDE</h3>

Increase in temperature will increase the solubility of the solid.
By adding NaF - decrease the solubility of BaF₂
By adding HCl - increase the solubility of BaF₂

<h3>BARIUM FLUORIDE</h3>

It is colourless solid that occur as rare mineral
It is corroded by moisture
It is used in window of IR spectroscopy

Hence the barium fluoride the solubility decreases on adding NaF and increases on adding HCl.

Learn more about the solubility on

brainly.com/question/6841847

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3 0

2 years ago

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ

Ahat [919]

<u>Answer:</u> The Gibbs free energy of the reaction is 21.32 kJ/mol

<u>Explanation:</u>

The chemical equation follows:

$\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $37^oC=[273+37]K=310K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}$

$\text{[Oxaloacetate]}=0.130mM$

$[NADH]=2.0\times 10^2mM$

$\text{[Malate]}=1.37mM$

$[NAD^+]=490mM$

Putting values in above expression, we get:

$\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol$

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

4 0

3 years ago