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abruzzese [7]
3 years ago
7

Raising 100 grams of water from 40 to 60 °c (the specific heat capacity of water is 1 cal/g) requires the addition of What? A) 1

20 calories B) 1500 calories C) 2000 calories D) 2400 calories
Chemistry
2 answers:
NNADVOKAT [17]3 years ago
8 0
<h2><em>the answer to this question lady's is C</em></h2>
marissa [1.9K]3 years ago
3 0
Q=mc(delta T)
= (100 g)(1 cal/g(K))(60-40)
=100(20)
=2000 calories
letter C
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The tabulated data were collected for this reaction:
erastovalidia [21]

Answer:

ai) Rate law,  Rate = k [CH_3 Cl] [Cl_2]^{0.5}

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)

If A + B \rightarrow C + D, the rate of backward reaction is given by:  

Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1

Using the information provided in lines 3 and 4 of the table and insering the value of a:

0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\

0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5

The rate law is: Rate = k [CH_3 Cl] [Cl_2]^{0.5}

The rate constant k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}} then becomes:

k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25

b) Overall order of reaction =  a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0
3 years ago
The most common fuel for nuclear fission reactors is:
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ra1l [238]

Answer:

A option is correct.

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Explanation:

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Answer:

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During an experiment, 321.1 grams of strontium phosphite are measured into a beaker. How many moles of strontium phosphite are i
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321.1 g of strontium phosphite will contain 0.76 moles of strontium phosphite in the beaker.

<u>Explanation:</u>

It is known that 1 mole of any substance is equal to the molar mass of that substance. So for the case of strontium phosphite, the molar mass is found to be 420.8 g/mol.

This means 1 mole of Strontium phosphite will contain 420.8 g of strontium phosphite.

Then 1 g of strontium phosphite will have \frac{1}{420.8} moles of strontium phosphite.

So, 321.1 g of strontium phosphite will contain \frac{321.1}{420.8}=0.76 moles

Thus, 321.1 g of strontium phosphite will contain 0.76 moles of strontium phosphite in the beaker.

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