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4 years ago

What is meant by Critical temperatue?

Chemistry

1 answer:

yuradex [85]4 years ago

4 0

Answer:

The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.

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As frequency is increased, wavelength … <br> finish the sentence

pickupchik [31]

The wave length decreases

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3 years ago

The specific heat of aluminum is 0.897 J/g•°C. Which equation would you use to calculate the amount of heat needed to raise the

docker41 [41]

Q = 0.75 g x 0.897 J/g•°C x 22°C

6 0

3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

3 years ago

What is the Bo element plz help

Effectus [21]

There is no Bo element, however there is a B element which is Boron

8 0

3 years ago

Read 2 more answers

The osmotic pressure of a solution formed by dissolving 35.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25°c is __________ at

Lunna [17]

Mass of aspirin = 0.025 g
Molar mass of C9H8O4 is 180.1583 g/mol
moles of aspirin = .025g / 180.1583 g/mol = 0.000138767 moles
volume solution = .250 L
molarity of the solution = 0.000138767 moles / .250L =5.551 x 10 ^-04 Moles / liter
for aspirin i = Vant'Hoff factor = 1 particle in solution
T = 25 + 273 =298 K
osmotic pressure = M x R x T x i =
5.551 x 10 ^-04 mole L -1 x 0.08206 L atm K−1 mol−1 x 298 K x 1 = 0.0136 atmospheres

6 0

3 years ago

Read 2 more answers