All categories

2 years ago

What would happen to the pressure if the volume were reduced to 0.5 l and the temperature increased to 260 ∘c?

1 answer:

8 0

<span>The ideal gas law. PV=nRT pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273) Original data Pressure 1 atmosphere Volume 1 liter Temp 25C = 298K New data Volume 0.5 liter pressure X Temp 260C = 533K P1v1T1 = P2v2T2 plug and chug. (1)(1)(293) = (x)(0.5)(533) Solve for X, which is the new pressure. </span>

You might be interested in

What is the name of the compound (NH4)2SO4

GREYUIT [131]

The answer is Ammonium sulfate

3 0

2 years ago

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

2 years ago

What are salt ? How they are used

Leno4ka [110]

There is different types of salt.
1)table salt
2)lime salt

7 0

3 years ago

3) _Na +<br> |H,80_ - _Na,304<br> _Na2SO4 + _H2

erik [133]

Hey

How was your day

3 0

2 years ago

The ka of hypochlorous acid (hclo) is 3.0 ⋅ 10−8 at 25.0 °c. calculate the ph of a 0.0375m hypochlorous acid solution.

Scrat [10]

We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375    0 0
Change    -x     +x +x
Equilibrium 0.0375-x    x x

We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)

Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47

6 0

3 years ago