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ELEN [110]
3 years ago
5

You mix 35mL of 0.54M NaI with 14mL of 0.84M Pb(NO3)2. How many grams of the precipitate will be created?

Chemistry
1 answer:
svlad2 [7]3 years ago
8 0
You mix 35mL of 0.54M NaI with 14mL of 0.84M Pb(NO3)2.
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A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each com
Fynjy0 [20]

Answer:

Mole fraction of CH_4O = 0.58

Mole fraction of C_2H_6O = 0.42

Explanation:

Let the mass of CH_4O and C_2H_6O = x g

Molar mass of CH_4O = 33.035 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}

Moles_{CH_4O}=\frac{x}{33.035}\ mol

Molar mass of C_2H_6O = 46.07 g/mol

Thus,

Moles= \frac{x\ g}{46.07\ g/mol}

Moles_{C_2H_6O}=\frac{x}{46.07}\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}

Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58

Mole fraction of C_2H_6O = 1 - 0.58 = 0.42

6 0
3 years ago
Put the steps in the process of photosynthesis in order
shepuryov [24]

Step 1-Light Dependent

CO2 and H2O enter the leaf

Step 2- Light Dependent

Light hits the pigment in the membrane of a thylakoid, splitting the H2O into O2

Step 3- Light Dependent

The electrons move down to enzymes

Step 4-Light Dependent

Sunlight hits the second pigment molecule allowing the enzymes to convert ADP to ATP and NADP+ gets converted to NADPH

Step 5-Light independent

The ATP and NADPH is used by the calvin cycle as a power source for converting carbon dioxide from the atmosphere into simple sugar glucose.

Step 6-Light independent

The calvin cycle converts 3CO2 molecules from the atmosphere to glucose

calvin cycle

The second of two major stages in photosynthesis (following the light reactions), involving atmospheric CO2 fixation and reduction of the fixed carbon into carbohydrate.

4 0
3 years ago
A substance has a volume of 20mL and a density of 2.5 g/mL. what is it’s mass?
DochEvi [55]
2.5X20=50g
50g should be the right answer
Mass=volumeXdenisty.
4 0
3 years ago
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo
siniylev [52]

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

3 0
3 years ago
Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid? Acid pKa Hydrobromic Acid −5
siniylev [52]

Answer:

10^{-3.4

Explanation:

Let us first take a look at the image below;

In the acid - base  reaction; we can see the transfer of electrons that takes place;

We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.

The stronger acid is shown with the one with more negative P_{Ka} Value.

∴ The equilibrium constant for the acid-base reaction is expressed as:

K_{eq}= \frac{K_a of reactant acid}{K_a of product acid}

      = \frac{10^{-pK} of reactant acid}{10^{-pk} of product acid}

From  P_{Ka} Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.

From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:

K_{eq}=\frac{10^{-pK} of hydrobromic acid}{10^{-pk} of Ethyloxonium acid}

K_{eq}=\frac{10^{-5.8} of hydrobromic acid}{10^{-2.4} of Ethyloxonium acid}

       = 10^{-3.4}

6 0
3 years ago
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