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Lelechka [254]
3 years ago
6

Why is it important for you to learn where the safety equipment is located in the lab?

Chemistry
1 answer:
Mariana [72]3 years ago
6 0
It is really very important to know where the safety equipments are placed in lab because when we know it then accident wont occur in lab . it will also helps us to feel secure while doing any kinds of experiment
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What is the basic definition of carbohydrates
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Which of these common substances is a mixture?
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Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

1810.04571428[mL]=V_2

Adjusting for significant figures, this gives V_2=1810[mL]

4 0
2 years ago
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