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lions [1.4K]
3 years ago
11

According to Le Chatelier’s principle, what happens when the concentration of reactants is doubled ina chemical reaction that wa

s in equilibrium?
Chemistry
2 answers:
ale4655 [162]3 years ago
4 0

Answer:

C

Explanation:

Paraphin [41]3 years ago
3 0
The equilibrium shifts to the right to favor the formation of the products. Le Chatelier's principle states that a system will do what it takes to negate any changes made or added to it.
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What is the full meaning of DR.HERC.​
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Answer:

It is

D- <u><em>DEFINE</em></u> the problem

R-<em><u>RESEARCH</u></em> on the problem

H- Carry out a <u><em>HYPOTHESIS</em></u>

E- carry out an <em><u>EXPERIMENT</u></em>.

R-Analyse the <u><em>RESULT</em></u>

C-summarise the <u><em>CONCLUSION</em></u>.

Explanation:

Hope it helps.

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3 years ago
A sample of 25mls of HCL was mixed with 25mls of KOH of the same concentration in a calorimeter. As a result the temperature ros
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3 years ago
The hardest known mineral is _____. topaz talc diamond graphite
klemol [59]
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4 years ago
Read 2 more answers
If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

8 0
3 years ago
A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet
stira [4]

Explanation:

The given data is as follows.

           Mass of apple sauce mixture = 454 kg

           Heat added (Q) = 121300 kJ

 Heat capacity (C_{p}) of apple sauce at 32.8^{o}C = 4.0177 kJ/kg^{o}C

So, Heat given by heat exchanger = heat taken by apple sauce

                            Q = mC_{p} \Delta T

or,                    Q = mC_{p} (T_{f} - T_{i})  

Putting the given values into the above formula as follows.

                     Q = mC_{p} (T_{f} - T_{i})  

              121300 kJ = 454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)

                      T_{f} = 76.5^{o}C

Thus, we can conclude that outlet temperature of the apple sauce is 76.5^{o}C.

3 0
4 years ago
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