what is the solubility of salt at 60°c if 2gm of salt remains as a residue when 7gm of saturated solution prepared at 60°c is dired?​

Answers

Answer 1

The first thing that was a good app and it would have to come back and forth to me when it comes out of my account to get my own account to the app to be able to use my account to make my account and my tablet and my phone number is not working correctly

Explanation:


Related Questions

Which starting circuit uses fuses, switches, and smaller wires to energize a relay and solenoid?

Starter Main Circuit
Starter Control Circuit
Starter Ground Path
Battery Positive Cable

Answers

Answer:

Option B (Starter Control Circuit) is the right option.

Explanation:

This same switching is normally put upon this isolated side of something like the transmission Arduino microcontroller throughout the configuration that is using the ignition just to command the broadcast.It uses a secondary relay isolated to regulate electrical current throughout the solenoid starting system.

All other given options are not related to the given instance. So the above option is correct.

3. Measurement ofla dimension
is
written as a

Answers

Answer:

Dimensions are expressed as width by length by height or depth in three-dimensional space.

hope it helps

A key joint in a precision machining process has a lower specification limit of a width of 0.99 mm and an upper specification limit of 1.01 mm. The standard deviation is 0.005 mm and the mean is 1 mm. What is the process capability index for the precision machining process

Answers

Answer:

the process capability index for the precision machining process is 0.667

Explanation:

Given the data in the question;

Lower Specification Limits LSL = 0.99 mm

Upper Specification Limits USL = 1.01 mm.

The standard deviation σ = 0.005 mm

mean μ = 1 mm

capability index Cpk = ?

Cpk = min( USL - μ / 3σ, μ - LSL / 3σ )

we substitute

Cpk = min( 1.01 mm - 1 mm / 3(0.005 mm ), 1 - 0.99 mm / 3(0.005 mm) )

Cpk = min( 0.01 mm / 0.015 mm, 0.01 mm / 0.015 mm )

Cpk = min( 0.6666, 0.6666 )

∴ Cpk = 0.6666 ≈ 0.667

Therefore, the process capability index for the precision machining process is 0.667

describe an engineering advance that has led to greater knowledge about the solar system

Answers

development of the space shuttle to allow humans to explore the moon

Answer:

Space Rovers

Explanation:

The engineering advancement of the invention space rovers has changed space exploration tremendously. The first space rover ever to go out onto another planet was made by NASA and put on mars to learn more about the red planet and see if there where any plausible signs of past or maybe even present life. This mission was incredibly successful and for the first time in space exploration history, there was a spacecraft that you could move around that could with do the conditions of the planet for weeks, months, sometimes years at a time, moving around taking pictures and constantly collecting data. This was revolutionary for scientists, this way they could get copious amounts of information just on rover and not having to send multiple crafts only to die in the first few hours but sending back a relative amount of information. With rovers you can have them collect samples and bring the back to earth, not something  that could’ve been done without a human, until rovers. They have completely changed and improved collecting data from other planets and have been one of the most revolutionary inventions when it comes to space exploration.

This is what I put for the answer, hope it helps in some way.

Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint space of 5.4 mm (0.210 in.) is allowed between standard rails of length 11.9 m (39 ft), what is the highest possible temperature that can be tolerated without the introduction of thermal stresses

Answers

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

Determine the highest possible temperature

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l )  = ∝ * ΔT

                                    ( 5.4 * 10^-3 / 11.9 )  = 12.1 * 10^-6 * ( T2 - 4 )

∴  ( T2 - 4 ) =  ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

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