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Slav-nsk [51]
4 years ago
5

Which of the species below can act as a lewis base? bf3 nf3 oh− alcl3?

Chemistry
1 answer:
IRINA_888 [86]4 years ago
6 0
Among BF₃, NF₃, OH⁻ and ALCl₃ both NF₃ and OH⁻ acts as a lewis base.

Explaination:
According to lewis theory of acid and base, base is a specie which has either lone pair of electrons or is negatively charged. So, in the given compounds Nitrogen trifluoride contain lone pair of electron on nitrogen and Hydroxide contains a negative charge on oxygen. And botha are capable to donate these electrons to a lewis acid (electrophiles).

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28 ml of 0.10 m hcl is added to 60 ml of 0.10 m sr(oh)2. determine the concentration of oh− in the resulting solution.
Blababa [14]
Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]

Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess. 
Follow the steps as

1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.

n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-

Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess

n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.

Total volume= V acid + V base= 28 ml + 60 ml = 98 ml

Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M

The answer is 0.009 M.


5 0
3 years ago
Read 2 more answers
Question 116 g of sunflower oil is
Luba_88 [7]
<h2>Answer:  125.41 mL</h2>

Explanation:

Volume = mass ÷ density

             =  116 g  ÷  0.925 g/mL

             =  125.41 mL

<h3>A 116 g of sunflower oil of 0.925 g/mL has a volume of 125.41 mL.</h3>
7 0
3 years ago
You wish to construct a buffer of pH=7.0. Which of the following weak acids (w/ corresponding conjugate base) would you select?
Alex17521 [72]

Answer:

C.) HOCl Ka=3.5x10^-8

Explanation:

In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below

we Know that

pKa= -log(Ka)

therefore

A) pKa of  HClO2 = -log(1.2 x 10^-2)

=1.9208

B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644

C)  pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45

D) pKa of HCN = -log(4 x 1 0^-10)=  9.3979

If we consider the  Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution

The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.

So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.

Hence, HOCl will be chosen for buffer construction.

3 0
3 years ago
Some people say that only gay and bisexual men are likely to be infected with hub true false
shusha [124]

Answer:

They do have a larger chance, but others might be infected too.

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3 0
3 years ago
Read 2 more answers
A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w
Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)

c=5760J/^oC

\Delta T=0.570^oC

To calculate the enthalpy change, we use the formula:

\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = \frac{3283.2J}{0.1326g}=24760.181J/g

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol

4 0
3 years ago
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