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Inessa05 [86]
3 years ago
9

A compound which is mineral acid is generally classified as a????? urgent please​

Chemistry
2 answers:
kobusy [5.1K]3 years ago
7 0

hydrochloric acid

that's the answer

andrey2020 [161]3 years ago
3 0

Answer: An inorganic acid.

Explanation: An inorganic acid (also called a mineral acid) is an acid derived from one or more inorganic compounds. All inorganic acids form hydrogen ions and the conjugate base ions when dissolved in water. Commonly used inorganic acids are sulfuric acid (H2SO4), hydrochloric acid (HCl), and nitric acid (HNO3). Hope this helps :)

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Position (m)
Alex777 [14]

Option C. The object is returning to the start at a constant speed.

<h3>Data points of the Position vs Time graph</h3>

The following data points will be used to determine the motion of the object.

<u>Position               Time</u>

12                          4

10                          6

2                            8

0                           10

From the data above, the position of the object is decreasing towards zero or start point.

Thus, the object is returning to the start at a constant speed.

Learn more about position here: brainly.com/question/2364404

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8 0
2 years ago
The factor that is changed throughout an experiment is called the _______. A. apparatus B. constant C. variable D. hypothesis
ser-zykov [4K]
The variable is what changes during an experiment. Hopefully this helped! :)
4 0
3 years ago
Read 2 more answers
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
The reaction type represented by AB ---&gt; A + B is known as
nordsb [41]

Answer:

A. synthesis

Explanation:

8 0
3 years ago
Calculate the cfse for a d^3 system in an octahedral field in units of ∆_o. In other words, do not enter "∆_o" with your answer.
Rudik [331]

Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q

To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.

[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.

The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.

To learn more about crystal field stabilization energy  visit:brainly.com/question/29389010

#SPJ4

8 0
1 year ago
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