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OLga [1]
3 years ago
9

Which statement is true

Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
7 0

Answer:

a

Explanation:

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What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B
natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

A = A_{0}e^{-kt}

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

ln[A] = ln[A]_{0}-kt

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

7 0
3 years ago
A 35,000 Newton car runs up a hill that is 45.5 meters high in 1.6 seconds. what power did it exert?
Evgesh-ka [11]

Answer:

995.313KW

Explanation:

the explanation is in the picture

please like and Mark as brainliest

7 0
2 years ago
6. Three ice cubes are placed inside a glass of hot chocolate. Which of the following best
ycow [4]
It’s the first 1 M yea it’s the first one
4 0
2 years ago
What is the molarity of a solution containing 55.8 g of mgcl2 dissolved in 1.00 l of solution?
MakcuM [25]
The answer is 0.59 M.

Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l

So, 1 mol has 95.2 g/l.

Our solution contains 55.8g in 1 l  of solution, which is 55.8 g/l

Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
7 0
3 years ago
Read 2 more answers
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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