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-Dominant- [34]
3 years ago
5

How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26ÁC? Show all work used to solve this pr

oblem
Chemistry
1 answer:
klio [65]3 years ago
6 0
The answer is 7.33 g.

<span>To calculate this, we will use the the ideal gas law:
PV = nRT
where
P - pressure of the gas,
V - volume of the gas,
n - amount of substance of gas,
R - gas constant,
T - temperature of the gas.</span>

Since the amount of substance of gas (n) can be expressed as mass (m) divided by molar mass (M), then:

PV = RTm/M

It is given:

P = 0.98 atm

V = 10.2 l

T = 26°C = 299.15 K 

R = 0.082 l atm/Kmol (gas constant)

M (H2O) = 2Ar(H) + Ar(O) = 2*1 + 16 = 2 + 16 = 18g

m = ?

Since PV = RTm/M, then:

m = PVM/RT

m = 0.98 · 10.2 · 18 / 0.082 · 299.15 = 179.928/24.5303 = 7.33 g

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.
Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. n_{CuO}= \frac{4}{80}= 0,05 (mol)

⇒n_{CuCl2}= n_{CuO}=0,05 mol

⇒m_{CuCl2}= 0,05×135=6,75 (g)

c. n_{HCl}=2× n_{CuO}=0,1 (mol)

⇒m_{HCl}= 0,1×36,5= 3,65 (g)

⇒m_{dd HCl}= \frac{m_{HCl}}{10}×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }×100=\frac{6,75}{36,5+4} ×100≈ 16,67%

8 0
3 years ago
why did mendeleev leave blank spaces on his periodic table? did later discoveries justify his predictions?
schepotkina [342]

Answer:

mendeleev left a space

Explanation:

so the periodic table can be organize

7 0
2 years ago
Which wave has a disturbance that is perpendicular to the wave motion?
Otrada [13]
It B I just took the test ;)
7 0
3 years ago
Read 2 more answers
2NaOH + H2SO4 &gt; Na2SO4 + 2H2O
Pepsi [2]

Answer:

To calculate the number we need molar mass.

6 0
3 years ago
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