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n200080 [17]
3 years ago
14

Classify the following as an element, compound, homogeneous mixture, or heterogeneous

Chemistry
1 answer:
Archy [21]3 years ago
3 0
The answer is D. Heterogeneous mixture
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Can someone answer question 3 ASAP please
allsm [11]

Answer:

Molecules must collide with sufficient energy, known as the activation energy, so that chemical bonds can break. Molecules must collide with the proper orientation. A collision that meets these two criteria, and that results in a chemical reaction, is known as a successful collision or an effective collision.

5 0
3 years ago
How many grams of Pb are contained in a mixture of 0.135 kg each of PbCl(OH) and Pb2Cl2CO3?
Dahasolnce [82]

Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg)  = 135 g

Molecular mass of PbCl(OH)  = 207+35.5+16+1  = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g  PbCl(OH)  :

(207 g Pb /  259.5 g PbClOH) * 135g PbClOH  =

0.79768 * 135 =>  107.68 g of Pb

For Pb2Cl2CO3  :

Given the mass of Pb2Cl2CO3  :

0.135 Kg = 0.135 Kgx(1000g / 1Kg)  = 135 g

Molecular mass of Pb2Cl2CO3  = 2*207+2*35.5+12+3*16  = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3  = 2*207 = 414 g

Hence mass of Pb in 135 g  Pb2Cl2CO3:

(414 g Pb /  545 g PbClOH) * 135g PbClOH  =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3


Hope that helps!

8 0
3 years ago
Read 2 more answers
How many moles of ammonia (nh3 will be produced from 4.0 moles of nitrogen (n2?
Lana71 [14]
Standard equation would be N2(g)+3H2(g)==>2NH3(g), so through stoichiometry, (4 mol N2)(2mol NH3/1 mol N2), assuming excess H2, would yield 8 moles of NH3.  
3 0
3 years ago
When 40 grams of ammonium nitrate explode 14 grams of nitrogen and 8 grams of oxygen form. How many grams of water form
DerKrebs [107]

Answer:

18 grams of water

Explanation:

The Balance Chemical Reaction is as follow,

                     2 NH₄NO₃    →    2 N₂  +  O₂  +  4 H₂O

According to Equation,

                     160 g (2 moles) NH₄NO₃ produces  =  72 g (4 moles) of H₂O

So,

                     40 g of NH₄NO₃ will produce  =  X g of H₂O

Solving for X,

                     X =  (40 g × 72 g) ÷ 160 g

                     X  =  18 g of H₂O

<em>Hope This Helps!</em>

4 0
2 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0
3 years ago
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