All categories

3 years ago

I need help! FAST!!!

Chemistry

1 answer:

aliina [53]3 years ago

5 0

Answer:

The second answer choice

Explanation:

X seems to list attributes for opaque objects, and y lists attributes of Transparent objects.

You might be interested in

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

How many grams of NaCl are in 2.0 Liters of a 3.0 M NaCl solution?

Naddik [55]

Answer:

350.64g

Explanation:

So first you must know that M is mol/L

Next solve the problem using dimensional analysis

2L NaCl (3 mol/L) = 6 mol NaCl

After you got the number of moles you should look at your periodic table to find the molar mass

I see that it's 58.44g/mol

Use dimensional analysis again!

6 mol (58.44g/mol) = 350.64g

Don't forget to make me brainliest!

4 0

3 years ago

Read the following chemical equation.

elena55 [62]

Answer:

Iron gains three electrons.

5 0

3 years ago

What problems could arise if scientists use different units to measure height

Shkiper50 [21]

Scientists can measure the height in different units but problem could arise when they compare all the measurements. That is the reason there is standard units for measurements.

There may be error arises when an American scientist is measuring the height of an object in inches and other Australian scientist is measuring the height of same object in meters. Their data cannot be compared because they are using different units to measure height.

5 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

I need help! FAST!!!​

I need help! FAST!!!