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sergeinik [125]
4 years ago
7

Please help!

Chemistry
1 answer:
lyudmila [28]4 years ago
4 0
<h3>The equation</h3>

4.94 x c . (Tm-23.6)=C.(23-6-21.9)+123.13 x c. (23.6-21.9)

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, ie the heat received/absorbed is the same as the heat released  

Q in = Q out  

Heat can be calculated using the formula:  

Q = mc∆T  

A calorimeter is a device used to measure the specific heat of material  

A metal is put into a calorimeter that contains water and there will be heat transfer:  

\tt \displaystyle m_mc_m (T_m-T)=m_wc_w(T-Tw)

m = metal  

w = water  

T = the final temperature of the mixture  

mass of metal =(Nickel) = 4.94 g

mass of calorimeter = 12.5 g

mass of water = 123.13 g (135.63 - 12.5)

The equation

Q released (metal) = Q absorbed(calorimeter+water)

  • Qmetal = 4.94 x c . (Tm-23.60)
  • Q calorimeter = C.(23-6-21.9) --> C = heat capacity of calorimeter
  • Q water = 123.13 x c. (23.6-21.9)

The equation :

4.94 x c . (Tm-23.6)=C.(23-6-21.9)+123.13 x c. (23.6-21.9)

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What volume of hydrogen will be produced at STP by the reaction of 78.33 g of aluminum with excess water
nydimaria [60]
The balanced equation for the above reaction is 
2Al + 6H₂O ---> 2Al(OH)₃ + 3H₂
stoichiometry of Al to H₂ is 2:3
number of Al moles reacted - 78.33 g / 27 g/mol = 2.901 mol
according to molar ratio 
2 mol of Al forms - 3 mol of H₂
therefore 2.901 mol of Al - forms 3/2 x 2.901 = 4.352 mol

molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP
if 1 mol occupies 22.4 L
then 4.352 mol occupies - 22.4 L/mol x 4.352 mol = 97.48 L
volume occupied by H₂ is 97.48 L
4 0
3 years ago
Convert 26.02 x 1023 molecules of C2H8 to grams. Round your answer to the hundredths place.
aliina [53]

Answer:

x= 138.24 g

Explanation:

We use the avogradro's number

6.023 x 10^23 molecules -> 1 mol C2H8

26.02 x 10^23 molecules -> x

x= (26.02 x 10^23 molecules  * 1 mol C2H8 )/6.023 x 10^23 molecules

x= 4.32 mol C2H8

1 mol C2H8     -> 32 g

4.32 mol C2H8 -> x

x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8

x= 138.24 g

4 0
4 years ago
How many atoms are in 20.0g Ca?
Vinil7 [7]

Answer: 6.022 x10^23 atoms

Explanation:

8 0
4 years ago
Can anyone check my chemistry answers?
frutty [35]
I check all of them. most of them correct but one

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4 0
4 years ago
The initial reaction rate for the elementary reaction 2A + B → 4C was measured as a function of temperature when the concentrati
Artist 52 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776\frac{kJ}{mole}

b) The frequency factor is 1.77 ×10^{18}

c) The rate constant is 0.00033 (\frac{dm^{3} }{mole} )^{2}\frac{1}{s}

Explanation:

From the question the elementary reaction for A and B is given as

      2A + B → 4C

The rate equation the elementary reaction is

      -r_{A} = k[A]^{2}[B]

            =  k[2]^{2}[1.5]

            = 6k

     k = \frac{-r_{A} }{6}

      When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

                    k = Aexp(-\frac{E_{a} }{RT} )

             taking ln of both sides we have

                   lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}

        Considering the graph for the rate constant ln k and (\frac{1}{T} ) the slope from the equation is -(\frac{E_{a} }{R}) and the intercept is ln A

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of ln k  vs (\frac{1}{T} )  is shown on the fourth uploaded image

  From the graph we can see that the slope is -(\frac{E_{a} }{R} ) = - 15008

Now we can obtain the activation energy E_{a} by making it the subject in the equation also generally R which is the gas constant is 8.145 \frac{J}{kmole}

                E_{a}  = 15008 ×  8,3145\frac{J}{molK}  

                     = 124\frac{KJ}{mole}

    Hence the activation energy is = 124\frac{KJ}{mole}

b) From the graph its intercept is ln A = 42.019

                                                          A = exp(42.019)

                                                             =1.77 × 10^{18}

Hence the frquency factor A is  =1.77 × 10^{18}

c) From the equation of rate constant

                                          lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}

We have

                ln k = 42.019 - 15008 * (\frac{1}{300} )

                      k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}

Hence the rate constant is k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}    

6 0
3 years ago
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