2. A destructive, rotating column of air that has very high wind speeds and that is sometimes visible

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

3 0

3 years ago

Choose the nonmetallic elements from the list. Check all that apply yttrium: oxygen: boron: polonium: argon: gallium: carbon:

BaLLatris [955]

Answer:

B, E, & G

Oxygen, Argon, & Carbon

7 0

3 years ago

Read 2 more answers

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

3 years ago

Determine which compound would raise the boiling point of water the most

umka2103 [35]

Because all the compounds are at the same concentration, the one that can produce more particles in solution will be the one that will raise the boiling point the most.

A. 2.0 M (NH4)3PO4 will produce 4 particles per molecule formula

8 0

3 years ago