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3 years ago

What are salt ? How they are used

Chemistry

1 answer:

Leno4ka [110]3 years ago

7 0

There is different types of salt.
1)table salt
2)lime salt

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A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop in 7.5 seconds? What is the cars acc

Marianna [84]

Answer:

The acceleration is: $4m/s^2$

Explanation:

Given

$u = 30.0m/s$ --- The initial velocity

$t = 7.5s$ --- time

$v = 0m/s$ -- The final velocity

Required

Determine the acceleration

To do this, we make use of the first equation of motion

$v = u - at$

We used negative because the car was coming to stop.

This gives:

$0 = 30 - 7.5 * a$

$0 = 30 - 7.5a$

Collect like terms

$7.5a = 30$

Solve for a

$a = \frac{30}{7.5}$

$a = 4m/s^2$

6 0

3 years ago

What are the relationships between melting and freezing of a mixture

Kaylis [27]

The temperature is the same but the heat flow is the opposite.

4 0

2 years ago

Give an example of instantaneous speed

OleMash [197]

Cheetah who is running with speed of 80 miles per hour then it is his instantaneous speed becaue i shows as in per hour speed..... hope it helps

3 0

2 years ago

Read 2 more answers

Carl crumbled a piece of paper before throwing it in the trash can. Which is true about the paper after it has been crumbled

tiny-mole [99]

Answer is <span>C: It's appearance changed.</span>

4 0

2 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

2 years ago