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Galina-37 [17]
4 years ago
10

You are performing a chemical reaction in a test tube. The test tube gets colder as the reaction takes place. This chemical reac

tion is... O a.a double replacement reaction. O b.endothermic Oc.a single replacement reaction. O d.a decomposition reaction. e. exothermic
Chemistry
1 answer:
Liula [17]4 years ago
4 0

Answer: Option (b) is the correct answer.

Explanation:

A chemical reaction in which heat energy is absorbed by the reactant molecules is known as an endothermic reaction.

Therefore, upon completion of this reaction the container in which reaction is carried out becomes colder.

A chemical reaction in which heat energy is released by the reactant molecules is known as an exothermic reaction.

Therefore, upon completion or during this type of reaction the container in which reaction is carried out becomes hot.

Thus, we can conclude that when the test tube gets colder as the reaction takes place then it means this chemical reaction is endothermic reaction.

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A Cu/Cu2 concentration cell has a voltage of 0.22 V at 25 o C. The concentration of Cu2 in one of the half-cells is 1.5 x 10-3 M
Olin [163]

Answer:

The concentration is  [Cu^{2+}]_a  = 10^{-10.269}  

Explanation:

From the question we are told that

    The voltage of the cell is  E =  0.22 \  V

   

   

Generally the reaction at the cathode is  

  Cu^{2+} _{(aq)} + 2e^{-} \to  Cu_{s} the half cell voltage is  V_c =  0.337 V

Generally the reaction at the anode is    

   Cu _{(s)} \to Cu^{2+} _{(aq)} +  2e^{-}  the half cell voltage is  V_a = -0.337 V

Gnerally the reaction of the cell is  

    Cu_{(s)} + Cu^{2+} _{(aq)} \to Cu^{2+}_{(aq)} +  Cu_{(s)}

At initial the voltage is  V  =  0 V

Generally the voltage of the cell at 25°C is  

       E =  V  - \frac{0.0591}{n} log \frac{[Cu^{2+}] _a}{[Cu^{2+}]_c}

Here n is number of  of electron and it is 2

So from the question we are told that one cell has a concentration 1.5 x 10-3 M

Let assume it is  [Cu^{2+}]_c

So

      0.22=  0  - \frac{0.0591}{2} log \frac{[Cu^{2+}] _a}{  1.5 * 10^{-3} }

=>    -7.445 =     log \frac{[Cu^{2+}] _a}{  1.5 * 10^{-3} }

=>  -7.445 =     log [Cu^{2+}_a] - log [1.5*10^{-3}]

=>    -7.445  + log [1.5*10^{-3} =     log [Cu^{2+}_a]

=>    -7.445  - 2.824 =     log [Cu^{2+}_a]

Taking the antilog

=>    [Cu^{2+}]_a  = 10^{-10.269}    

=>     [Cu^{2+}]_a  = 5.38 *10^{-11} \  M  

4 0
3 years ago
38 POINTS - PEER COUNSELING 2 QUESTION
brilliants [131]

anything except for a and d

7 0
3 years ago
Read 2 more answers
Given KNO3 + CuSO4 is a double replacement reaction, predict the products and then determine whether you would expect to see a p
slava [35]

Answer:

Products will be K2SO4 and Cu(NO3)2. A precipitate will not form.

Explanation:

Double replacement reactions are the switching of cations between two compounds. So in this case, the K+ and Cu 2+ cations will switch places.

KNO3 + CuSO4 -> K2SO4 + Cu(NO3)2

According to solubility rules, any substances containing NO3 or a Group 1 ion (K+ in this case) will be definitely soluble. There are a few exceptions to the Group 1 rule but it does not apply here. Potassium sulfate (K2SO4) contains a Group 1 ion so it will be soluble and not be a precipitate. And copper nitrate (Cu(NO3)2) contains nitrate so it will also be soluble and not be a precipitate.

4 0
3 years ago
Determine the location of the last significant place value by placing a bar over the digit.
Vlad1618 [11]

Answer:

8,040  

0.0300  

699.5  

2.000 x 102

0.90100  

90, 100  

4.7 x 10-8  

10,800,000.0  

3.01 x 1021

0.000410

Explanation:

First remember the following rules of determining the last significant place value :

1. The digits from 1-9 are all significant and zeros between significant digits are also significant.

2.  The trailing or ending zeroes are significant only in case of a decimal number otherwise they are ignored. However starting zeroes of such a number are not significant.

Now observing above rules, lets determine the location of the last significant place value of each given example. I am determining the location by turning the last significant place to bold.

1) 8,040

8,040

Location of the last significant place value is 3 and bar is over last significant digit that is 4. Number is not decimal so ending zero is ignored. Every non zero digit is a significant.

2)  0.0300

0.0300

Location is 3 and bar is over 0. Number has a decimal point so ending zero is not ignored but starting zeroes are ignored.

3) 699.5

699.5

Location is 4 and bar is over 5.

4) 2.000 x 10²

2.000 x 10²

Location is 4 and bar is over 0. This is because the number is decimal so trailing zeroes cannot be ignored. Also if we convert this number it becomes:

200.0 so last significant digit is 0 and location of last significant digit is 4.

5) 0.90100

0.90100

Location is 5 and bar is over 0. This is because in a number with decimal point starting zeroes are ignored but trailing zeroes after decimal point are not ignored. So we count from 9 and last significant digit is 0.

6) 90, 100

90, 100

Location is 3 and bar is over 1. This is because it is not a number with decimal point. So the trailing zeroes are ignored. The count starts from 9 and last significant is 1.

7) 4.7 x 10⁻⁸

4.7 x 10⁻⁸

Location is 2 and bar is over 7. This is because the starting zeroes in a number with a decimal point are ignored. So the first digit considered is 4 and last significant digit is 7. If we expand this number:

4.7 x 10⁻⁸ =    0.000000047 = 0.000000047

Here the starting zeroes are ignored because there is a decimal point in the number.

8) 10,800,000.0

10,800,000.0

Location is 9 and bar is over 0.  Number has a decimal point so ending zero is not ignored and last significant figure is 0.

However if the number is like:

10,800,000. Then location would be 8 and bar is over 0.

9) 3.01 x 10²¹

3.01 x 10²¹

Location is 3 and bar is over 1. Lets expand this number first

3.01 x 10²¹ = 3.01 x 1000000000000000000000

                  = 3010000000000000000000

So this is the number:

3010000000000000000000

Since this is number does not have a decimal point so the trailing zeroes are ignored. Hence the count starts from 3 and the last significant figure is 1

10) 0.000410

0.000410

Location is 3 and bar is over 0. This is because the number has a decimal point so the ending zero is not ignored but the starting zeroes are ignored according to the rules given above. Hence the first significant figure is 4 and last significant figure is 0.

6 0
3 years ago
A 4.000 g sample of an unknown metal, M, was completely burned in excess O2 to yield 0.02225 mol of the metal oxide, M2O3. What
Solnce55 [7]
4X + 3O₂ = 2X₂O₃

n(X₂O₃)=0.02225 mol
m(X)=4.000 g
x - the molar mass of metal

m(X)/4x=n(X₂O₃)/2

x=m(X)/{2n(X₂O₃)}

x=4.000/{2*0.02225)=89.89 g/mol

X=Y (yttrium)



7 0
4 years ago
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