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denis23 [38]
2 years ago
15

A change in temperature causes a change in density. \ True False

Chemistry
2 answers:
Ray Of Light [21]2 years ago
5 0

Answer:

it will be true

Explanation:

stealth61 [152]2 years ago
4 0
Yes it does change because the volume changes wi the temperature
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Consider the equation: 2NO(0) - N.04(). Using ONLY the information given by the equation which of the following
ratelena [41]

Answer:

By increasing the pressure, the molar concentration of  N2O4 will increase

Explanation:

We have the equation 2NO2 ⇔ N2O4

This equation is reversible and exotherm. By <u>decreasing the temperature</u>, the reaction will produce more energy, so the reaction will move to the right.  But a lower temperature also lowers the rate of the process, so, the temperature is set at a compromise value that allows N2O4 to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable

So <u>increasing the temperature</u> will shift the equilibrium to the left. The equilibrium shifts in the direction that consumes energy.

If we d<u>ecrease the concentration of NO2</u>, the equilibrium will shift to the left, resulting in forming more reactants.

To increase the molar concentration of the product N2O4, we have to <u>increase the pressure</u> of the system.

NO2 takes up more space than N2O4, so increasing the pressure would allow the reactant to collide more form more product.

By increasing the pressure, the molar concentration of  N2O4 will increase

7 0
3 years ago
Read 2 more answers
Calcium carbide, CaC2, is manufactured by reducing lime with carbon at high temperature. (The carbide is used in turn to make ac
sweet-ann [11.9K]

The energy of the carbide released is 7262.5MJ.

<h3>What is the energy?</h3>

We know that the reaction between calcium  oxide and carbon occurs in accordance with the reaction; CaO(s)+3C(s)----- > CaC_{2} (s)+CO(g). The reaction is seen to produce 464.8kJ of energy per mole of carbide produced.

Number of moles of CaC_{2} produced = 1000 * 10^3 g/64 g/mol

= 15625 moles of calcium carbide

If 1 mole of CaC_{2} transfers  464.8 * 10^3 J

15625 moles of calcium carbide transfers 15625 moles  * 464.8 * 10^3 J/ 1 mol

= 7262.5MJ

Learn more about reaction enthalpy:brainly.com/question/1657608

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7 0
1 year ago
Use the following image to answer the following questions. A student was performing an experiment using two identical cups made
Anettt [7]

Answer:

The water will eventually become the same temature (A)

3 0
2 years ago
Help me out with these three formulas
ddd [48]
Carbon dioxide it should be water oxygen
7 0
3 years ago
In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

4 0
3 years ago
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