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Natali [406]
3 years ago
6

HELP ASAP MY LESSON IS BEING TIMED!!!

Chemistry
2 answers:
77julia77 [94]3 years ago
7 0

Answer:

1. Newton's third law of motion

2. Unbalanced

3. Action

Ugo [173]3 years ago
7 0
It’s action for number 3
You might be interested in
Al+Mn2+>Al3++Mn. How many electrons will be lost in all?
Alisiya [41]
It is probably 6 just divide
5 0
3 years ago
This question is "Which procedure would best increase the amount of heat energy that is actually gained by calorimeter B? (1)inc
professor190 [17]
3 as long as it is warm or hot air.
3 0
3 years ago
the molar enthalpy of formation of carbondioxide is -393kjmol. calculate the heat released by the burning of 0.327g of carbon to
Arte-miy333 [17]

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

<u>Explanation</u>:

The standard enthalpy change of reaction,  Δ H ∘ , is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.

                                    C (s]  +  O 2(g] → CO 2(g]

Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.

First to convert  grams of carbon into moles,

use carbon's molar mass(12.011 g).

                    Moles of C = mass in gram / molar mass

                                        = 0.327 g  / 12.011 g

                     Moles of C = 0.027 moles

Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.

                                        =  0.027 moles C  \times 393 kJ

             Heat released  = 10.611  kJ.

So, when  0.027  moles of carbon react with enough oxygen gas, the reaction will give off  10.611 kJ  of heat.

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

7 0
3 years ago
Predict the sign of the entropy change of the system for each of the following reactions.
storchak [24]

Answer :

(a) The entropy change will also decreases.

(b) The entropy change will also increases.

(c) The entropy change will also decreases.

(d) The entropy change will also increases.

Explanation :

Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

As we are moving from solid state to liquid state to gaseous state, the entropy will be increases due to the increase in the disorderedness.

As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

(a) N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

In this reaction, the randomness of reactant molecules are less and as we move towards the formation of product the randomness become more that means the degree of disorderedness increase. So, the entropy change will also increases.

(c) 3C_2H_2(g)\rightarrow C_6H_6(g)

In this reaction, 3 mole of gaseous C_2H_2 react to give 1 mole of gaseous C_6H_6 that means randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(d) Al_2O_3(s)+3H_2(g)\rightarrow 2Al(s)+3H_2O(g)

In this reaction, the number of moles of gases are same on both side of the reaction but 1 mole of solid Al_2O_3 react to give 2 moles of solid aluminium that means randomness become more that means the degree of disorderedness increases. So, the entropy change will also increases.

5 0
3 years ago
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
5 0
3 years ago
Read 2 more answers
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