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fomenos
3 years ago
9

93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP

Chemistry
1 answer:
Dahasolnce [82]3 years ago
3 0
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
V= 93ml  to  liters =  93/1000= 0.093L
T=  10  +  273.15 = 283.15k

n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

At  STp    1  mole  =  22.4L
what about  3.72  x10^-3 moles

by  cross  multiplication

volume =  (3.72  x10^-3)mole  x  22.4L/  1  moles  = 0.083 L   or  83.3 Ml
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Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

        x              _________  25.2 units

x = 1.741x10⁻⁵mol

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C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

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3 years ago
TRUE FALSE? The MOLE is a unit in Chemistry that serves as a bridge between the ATOMIC and MACROSCOPIC worlds?​
Vanyuwa [196]

Answer:

True

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Hope this helps

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3 0
3 years ago
Consider the substances hydrogen (H2), fluorine (F2), and hydrogen fluoride (HF). Based on their molecular structures, how does
katovenus [111]

Answer:

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Explanation:

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3 years ago
If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat
sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated  solution contains the <u>maximum amount of a solute that will dissolve in a given  solvent at a specific temperature</u>.

An unsaturated solution contains <u>less solute than it  has the capacity to dissolve. </u>

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According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:

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b) 65 g in 100 mL of H₂O at 50 °C  ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0
3 years ago
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