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Sergeeva-Olga [200]
2 years ago
10

True or false when matter changes it’s state it is a physical change because it is the same substance the whole time.

Chemistry
1 answer:
Blababa [14]2 years ago
5 0

Answer:

true

Explanation:

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BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.
Yakvenalex [24]

Answer:

T_2=-125.58\°C

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Whereas solving for the final temperature T2, we get:

T_2=\frac{T_1P_2}{P_1}

Thus, we plug in the given data (temperature in Kelvins) to obtain:

T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C

Best regards!

3 0
2 years ago
Write the molecular formula in the of the following compounds.
tensa zangetsu [6.8K]
Calcium bicarbonate - Ca(HCO3)2
sodium peroxide - Na202
water - H20
silver nitrate - AgN03
potassium carbonate - K2CO3
sodium carbonate - Na2CO3
zinc chloride - ZnC12
calcium hydroxide - Ca(OH)2
magnesium chloride - MgC12
4 0
2 years ago
What is shown in the diagram below?
Colt1911 [192]
B I think is right. Hope this helps!
3 0
3 years ago
Read 2 more answers
If a basketball travels a distance of 4 meters in 5 seconds, what is its average speed?
creativ13 [48]

Answer:

i think d

Explanation:

7 0
3 years ago
One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo
frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J

Stape 2: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J

The total work is the sum of the two steps

W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J

4 0
3 years ago
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