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Dafna11 [192]
3 years ago
7

If a piece of space debris is too large to be a meteoroid and too small to be a planet, it could be _______.

Chemistry
2 answers:
butalik [34]3 years ago
3 0

I think it could be "an asteroid" .

otez555 [7]3 years ago
3 0
An asteroid or just another price of space junk if it’s too small to an asteroid and the other factors. Depends on the type of space junk and the size and if it actually is worth naming on the NASA computer. If it isn’t then, it’s junk. If it is, then you should name the asteroid on the NASA computer systems.
You might be interested in
. Which of the following statements best describes the relationship between elements and compounds?
vampirchik [111]
I think the answer is A
8 0
3 years ago
Which of the following reactions should have the larger emf under standard conditions? Why?
miskamm [114]

Answer:

Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)

Explanation:

If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.

The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.

On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.

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
3Al + 3 NH4ClO4 ---> Al2O3 + AlCl3 + 3 NO + 6H20
Tresset [83]

Answer:

9.63 L of NO

Explanation:

We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:

Mass of NH₄ClO₄ = 50 g

Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)

= 14 + 4 + 35.5 + 64

= 117.5 g/mol

Mole of NH₄ClO₄ =?

Mole = mass /molar mass

Mole of NH₄ClO₄ = 50/117.5

Mole of NH₄ClO₄ = 0.43 mole

Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:

3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O

From the balanced equation above,

3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.

Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.

Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:

1 mole of NO = 22.4 L

Therefore,

0.43 mole of NO = 0.43 × 22.4

0.43 mole of NO = 9.63 L

Thus, 9.63 L of NO were obtained from the reaction.

6 0
2 years ago
How many total moles of ions are released when the following sample dissolves completely in water? 6.188 x 10 21 formula units o
elixir [45]

Answer:

e. 3.08 x 10⁻² mol of ions.

Explanation:

  • Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).

  • We can get the no. of moles of NiCl₂ using cross multiplication:

1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.

??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.

∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.

  • NiCl₂ is ionized according to the equation:

NiCl₂ → Ni²⁺ + 2Cl⁻.

Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).

<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>

3 0
3 years ago
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