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BlackZzzverrR [31]
3 years ago
10

Calculate the mass of 1.35 moles of sodium chloride (NaCl).

Physics
1 answer:
Harlamova29_29 [7]3 years ago
4 0

Answer:d

Explanation:

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Un cuerpo suspendido en el aire es capaz de:
Rama09 [41]

Answer:

Explanation:

d

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3 years ago
Which simple machine is a doorknob?
Anestetic [448]

The answer is wheel and axle

3 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad
dexar [7]

Answer:

f = 4000 / 30 sec = 133.3    vibrations/sec

P = 1 / f = .0075 sec       period of 1 vibration

4 0
2 years ago
An aluminum wing on a passenger jet is 35 m long when its temperature is 17°C. At what temperature would the wing be 3 cm (0.03
Mnenie [13.5K]

Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

The linear expansion coefficient of aluminium \alpha =23.6\times 10^{-6}/^{\circ}C

We know that change in length is given by \Delta L=L\alpha \Delta T

So 0.03=35\times 23.6\times 10^{-6}\Delta T

\Delta T=36.32^{\circ}C

So final temperature =T_I+\Delta T=17+36.32=53.3196^{\circ}C

5 0
3 years ago
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