On May 3, Zirbal Corporation purchased 4,500 shares of its own stock for $36,000 cash. On November 4, Zirbal reissued 800 shares of this treasury stock for $7,200. Prepare the May 3 and November 4 journal entries to record Zirbal's purchase and reissuance of treasury stock.

Answers

Answer 1

Answer:

Explanation:

Each share costs $36000/4500

= $8

800 shares will cost 800*8

= $6400

Find attached the journal

On May 3, Zirbal Corporation Purchased 4,500 Shares Of Its Own Stock For $36,000 Cash. On November 4,

Related Questions

how many number of electrons are in aluminium and nitrogen in their free state?​

Answers

Answer:

MRCORRECT has answered the question

Explanation:

13 electrons

So... for the element of ALUMINUM, you already know that the atomic number tells you the number of electrons. That means there are 13 electrons in a aluminum atom. you can see there are two electrons in shell one, eight in shell two, and three in shell three.

7 electrons

So... for the element of NITROGEN, you already know that the atomic number tells you the number of electrons. That means there are 7 electrons in a nitrogen atom. Looking at the picture, you can see there are two electrons in shell one and five in shell two.

why is the international space station in orbit ​

Answers

Answer:

The ISS provides a location in the relative safety of low Earth orbit to test spacecraft systems that will be required for long-duration missions to the Moon and Mars.

Explanation:

what happens when light is not spread out

Answers

Answer:

Spreading out a wave over a larger area just causes the wave strength to weaken, but does not cause gaps to form. Therefore, if you look at photons as waves, spatial gaps never form in light as it travels through free space, no matter how dim it gets.Spreading out a wave over a larger area just causes the wave strength to weaken, but does not cause gaps to form. Therefore, if you look at photons as waves, spatial gaps never form in light as it travels through free space, no matter how dim it gets

Explanation:

what is the correct formula for the compound platinum (vi) flouride?

Answers

Answer: PtF6

Explanation:  is one of seventeen known binary hexafluorides.

What is the overall reduction potential for the reaction Agt (aq) + Cu(s)——>Ag(s) + Cu2+ (aq)?
-0.46 V
+0.46 V
+0.57 v
+1.14V

Answers

Answer:

it should be +0.46 :)

Explanation:

I just took the test and got it right.

Answer:

I think it's +1.14 V, but don't take my word for it..

Balance equation: _NO + _H2O = _NH3 + _O2

Answers

Answer:

4NO + 6H2O = 4NH3 + 5O2

Explanation:

Think about it this way: there are 4N's 4O's and 12H's and 10O's on both sides that's what makes it balanced

Hope this helps :)

Is the same reactant always the limiting reactant?

Answers

No because the limiting reactant is determined by calculating the amount of product therefore the limiting reactant will not be the same for a given reaction

mutations that neither benefit nor harm the organism have (blank) effect on the organisms survival

Answers

Answer:

I would say that the mutation has no effect on the organism, as it doesn't help or harm it.

hope this helps :)

Explanation:

Answer:

its a neutral

Explanation:

i did the test

How many grams of H20 will be formed when 32.0 g H2 reacts with 16.0 g
O2? *
2H2 + O2 + 2H2O
9.00 g
16.0 g
18.0 g
32.0g

Answers

Answer:

18.0 g

Explanation:

M(H2) = 2.0 g/mol

32.0g * 1 mol/2 g= 16 mol H2

M(O2) = 32 g/mol

16.0 g * 1mol/32 g= 0.5 mol O2

                                      2H2 + O2 ---> 2H2O

from reaction              2 mol    1 mol

given                         16 mol     0.5 mol

We see that O2 is a limiting reactant.

                                    2H2 + O2 +-------> 2H2O

from reaction                         1 mol            2 mol

given                                      0.5 mol        x mol

x =(0.5*2)/1= 1 mol H2O

M(H2O)= 18 g/mol

18 g/mol* 1 mol = 18 g H2O

Calculate the molar mass of ammonium chloride

Answers

Answer:

53.491 g/mol

Explanation:

Create the chemical compound and find each individual element's molar mass. Lastly, add them up.

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