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iVinArrow [24]
3 years ago
14

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Chemistry
2 answers:
iogann1982 [59]3 years ago
4 0

Answer:

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viktelen [127]3 years ago
3 0

Answer:

HmmmmmmmmmmI think the answer is 69

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Oxygen gas generated in an experiment is collected at 25.0°C in a bottle inverted in a trough of water. The total pressure is 1.
Nitella [24]

Answer:

0.007 mol

Explanation:

We can solve this problem using the ideal gas law:

PV = nRT

where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.

Keep in mind that when  we collect a gas over water we have to correct for the vapor pressure of water at  the temperature in the experiment.

Ptotal = PH₂O + PO₂  ⇒ PO₂ = Ptotal - PH₂O

Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.

P H₂O = 23.8 mmHg x 1 atm/760 mmHg =  0.031 atm

V = 1750 mL x 1 L/ 1000 mL = 0.175 L

T = (25 + 273) K = 298 K

PO₂ = 1 atm - 0.031 atm = 0.969 atm

n =  PV/RT = 0.969 atm x  0.1750 L / (0.08205 Latm/Kmol x 298 K)

n = 0.007 mol

6 0
2 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
2 years ago
Sodium dodecanoate is soluble in water even though it contains a large hydrophobic segment. How does this molecule interact with
velikii [3]

Answer:

Sodium laurate, also known as sodium dodecanoate, is a soap. It is the salt of lauric acid. It is an amphiphilic organic molecule which is composed of a hydrophilic head (polar ) and a hydrophobic tail (non-polar fatty acid).

In a aqueous solution, it leads to the formation of a micelle. The hydrophilic head of the molecule interacts with the surrounding polar solvent molecules. Thereby, making the micelle soluble in the solution. Whereas, the hydrophobic tails present in the core of micelle, interacts with the non-polar oil particles.

7 0
3 years ago
A student adds sugar to a cup of iced tea and a cup of hot tea. She notices that the time needed for the sugar to dissolve in ea
Natali [406]

Explanation:

the experiment conducted is the student adds sugar to a cup of iced tea and a cup of hot tea. She notices that the time needed for the sugar to dissolve in each cup is different. She thinks this has something to do with the temperature of the tea

hypothesis: If the student puts the sugar in both glasses of tea, then the sugar in the hot tea should dissolve quicker.

4 0
2 years ago
What is the concentration created when 380.0g of Pb(NO3)2 is added to 330.0 mL of solution?
Juliette [100K]

The concentration of lead nitrate is 3.48 M.

<u>Explanation:</u>

The molarity can be found by dividing moles of sucrose by its volume in litres. We can find the number of moles of sucrose by dividing the given mass by its molar mass. Now we can find the moles as,

Here mass of Pb(NO₃)₂ is 380 g

Molar mass of Pb(NO₃)₂ is 331.2 g/mol

Number of moles = $\frac{given mass}{molar mass}

                              =  $ \frac{380 g }{331.2 g/mol}

                              = 1.15 moles

Volume in Litres = 330 ml = 0.33 L

Molarity = $\frac{Moles}{Volume (L)}

          = 3.48 mol/L or 3.48 M

So the concentration of lead nitrate is 3.48 M.

7 0
2 years ago
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