The equation shows neutralization of an acid and a base to produce a salt and water

How many moles are in 2.3 grams of phosphorus? :(

tatiyna

Answer:

there are 0.074 moles in 2.3 grams of phosphorus

4 0

4 years ago

A shift of a cup of through

KatRina [158]

What are you asking?

5 0

3 years ago

Read 2 more answers

Are the properties of Rubidium more similar to those of cesium or those of strontium?

fenix001 [56]

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

Rubidium and Cesium are located in the first group on the periodic table.
Other elements in this group are lithium, sodium, potassium and francium
Strontium belongs to the second group on the periodic table.
The first group have a ns¹ valence shells electronic configuration.
They are all referred to as alkali metals

Learn more:

Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

#learnwithBrainly

5 0

3 years ago

What are 3 Properties of Convalent electrons

yKpoI14uk [10]

Do not ionize in solutions
Poor conductors of electricity/heat
Low melting/boiling points
gases or liquids at room temperature

4 0

3 years ago

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator

Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to 5.6

Explanation:

The equation for the reaction is :

$C_2H_5NH_2_(_a_q_) + H^+_(_a_q_) --- C_2H_5NH_{3(aq)}^+$

concentration of $C_2H_5NH_{2(aq)$ = 10%

10 g of $C_2H_5NH_{2(aq)$ in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of $C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL$

= 2.22 M

number of moles of $C_2H_5NH_{2(aq)$ in 20 mL can be determined as:

$= 20 mL \times 2.22 M= 44*10^{-3} mole$

Concentration of $C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}$

= 2.22 M

Similarly, The pKa Value of $C_2H_5NH_{2(aq)$ is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

$pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]$

$pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]$

$pH = 5.21$

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0

3 years ago