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podryga [215]
4 years ago
14

How might proper lab techniques impact a scientist work?

Chemistry
1 answer:
lana [24]4 years ago
7 0
Because if the the technique is wrong the scientist is wrong I’m sorry it’s a bad answer :(
You might be interested in
determine the percent yield for carbon dioxide if 4.50 moles of propane yielded 7.64 moles of carbon dioxide
ira [324]

Answer:

Percent yield = 57%

Explanation:

Given data:

Number of moles of propane = 4.50 mol

Number of moles of carbon dioxide = 7.64 mol

Percent yield = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Now we will compare the moles of propane and carbon dioxide.

                            C₃H₈            :            CO₂

                                 1               ;                3

                                  4.50        :              3×4.50 = 13.5 mol

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield =  7.64 mol / 13.5 mol × 100

Percent yield = 0.57× 100

Percent yield = 57%

3 0
3 years ago
What is the rate law for the reaction 2A + 2B + 2C --> products
-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
What type of reaction is Ni + MgSo4
kumpel [21]
Chemical- it produces ammonia.
4 0
3 years ago
Alkaline earth metals are very reactive because of how many valence electrons ?
Alborosie
Their is 2 valence electrons
8 0
3 years ago
Compared to ultraviolet light, an electromagnetic wave that has a higher frequency will also have ________. (4 points)
Jet001 [13]

Answer: shorter wavelength and equal speed

5 0
4 years ago
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