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2 years ago

What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 Pa to its container

1 answer:

3 0

Use the ideal gas law:

PV = nRT

so, T = PV / nR

n=0.5
V= 120 dm^3 = 120 L (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa = 0.147 atm (1 pa = 9.86 10^{-6} )

Put the values:

T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K

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When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is 2

tiny-mole [99]

Answer : The resulting concentration of $Na^+$ ion is, 4.5 M

Explanation : Given,

Concentration of $Na_2CO_3$ = $M_1$ = 3.0 M = 3.0 mol/L

Volume of $Na_2CO_3$ = $V_1$ = 70 mL = 0.07 L

Concentration of $NaHCO_3$ = $M_2$ = 1.0 M = 1.0 mol/L

Volume of $NaHCO_3$ = $V_2$ = 30 mL = 0.03 L

First we have to calculate the moles of $Na_2CO_3$ and $NaHCO_3$

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol$

and,

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol$

Now we have to calculate the moles of $Na^+$ ions.

As, 1 mole of $Na_2CO_3$ will give 2 moles of $Na^+$ ions

So, 0.21 moles of $Na_2CO_3$ will give $2\times 0.21=0.42$ moles of $Na^+$ ions

and,

As, 1 mole of $NaHCO_3$ will give 1 mole of $Na^+$ ions

So, 0.03 moles of $NaHCO_3$ will give 0.03 moles of $Na^+$ ions

So,

Total number of moles of $Na^+$ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of $Na^+$ ions.

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M$

Therefore, the resulting concentration of $Na^+$ ion is, 4.5 M

6 0

2 years ago

Which organism has cerebral ganglia in its head connected to ladder-like arrangements of nerve fibers?

Evgen [1.6K]

Your answer would be flatworms

6 0

2 years ago

Read 2 more answers

Explain how interactions between Earth’s spheres keep water cycling between them

Elenna [48]

There are several process in the nature, which keeps water cycling between them. By Precipitation, Earth's sphere gets water which we use in several ways, and by evaporation & sublimation, it goes back to upper spheres of the Earth's atmosphere and cycle happens again & again.

Hope this helps!

3 0

2 years ago

How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)? 2AgNO3 + CaC

nexus9112 [7]

Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction

2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).

From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂

Henceforth, 6.2 moles of AgNO₃ reacts with $\frac{6.2}{2}$ = 3.1 moles of CaCl₂.

1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-

A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.

C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.

D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃

Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.

6 0

2 years ago

an unknown metal of mass 512 g at a temperature of 15C is dropped into 325 g of water held in a 100 g aluminum container at the

ziro4ka [17]

Answer:

The specific heat capacity of the metal is 0.843J/g°C

Explanation:

Hello,

To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.

Heat gained by the metal = heat loss by water + calorimeter

Data,

Mass of metal (M1) = 512g

Mass of water (M2) = 325g

Initial temperature of the metal (T1) = 15°C

Initial temperature of water (T2) = 98°C

Final temperature of the mixture (T3) = 78°C

Specific heat capacity of metal (C1) = ?

Specific heat capacity of water (C2) = 4.184J/g°C

Heat loss = heat gain

M2C2(T2 - T3) = M1C1(T3 - T1)

325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)

1359.8 × 20 = 512C1 × 63

27196 = 32256C1

C1 = 27196 / 32256

C1 = 0.843J/g°C

The specific heat capacity of the metal is 0.843J/g°C

8 0

2 years ago