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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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Number of O atoms : 24

<h3>Further explanation</h3>

Given

C₆H₁₂O₆ compound

Required

Number of atoms

Solution

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6 0
3 years ago
Forming a hypothesis is accomplished through _______ reasoning.
vfiekz [6]
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Hope this helps!
4 0
3 years ago
Read 2 more answers
4) The initial rate of the reaction between substances P and Q was measured in a series of
ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

rate = k[P]^{2} [Q]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

k=\frac{rate}{[P]^{2}[Q] } \\  =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4

Second row:

2. Rate value:

rate =0.4* (0.10)^{2} * (0.10)\\\\        =4.0*10^-3mol.dm^-3.s^-1

3.Third row:

[Q]=\frac{rate}{k.[P]^{2} } \\     =9.6*10^-3 / (0.4 *(0.40)^{2} \\    =0.15mol.dm^{-3}

4. Fourth row:

[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}

6 0
3 years ago
What is the elevation of hachure line A?<br><br> 125 feet<br> 75 feet<br> 100 feet<br> 50 feet
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<span> elevation between index contours would be </span><span>125 feet</span>
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