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3 years ago

What volume of beaker contains exactly 2.23x10^-2 mol of nitrogen gas at STP?

Chemistry

1 answer:

nikklg [1K]3 years ago

4 0

Answer:

V = 0.5 L

Explanation:

Given data:

Moles of nitrogen = 2.23×10⁻² mol (0.0223 mol)

Temperature = 273 K

Pressure = 1 atm

Volume = ?

Solution:

PV = nRT

V = nRT / P

V = 0.0223 mol × 0.0821 atm. mol⁻¹. L . k⁻¹ × 273 K / 1 atm

V = 0.5 L

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You add 50.0 g of ice initially at ‒20.0 °C to 1.00 x 102 mL warm water at 67.0 °C. When all the ice melts, the water temperatur

xxTIMURxx [149]

Answer:

$T_2=17.8\°C$

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

$Q_{water}=-Q_{ice}$

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

$m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})$

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

$m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C$

Best regards.

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