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maks197457 [2]
3 years ago
7

Wood has a density of 0.531 g/cm3. What is the volume of 44 grams of wood?

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

<h3>The answer is 82.86 cm³</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass of wood = 44 g

density = 0.531 g/cm³

It's volume is

volume =  \frac{44}{0.531}  \\  = 82.86252354...

We have the final answer as

<h3>82.86 cm³</h3>

Hope this helps you

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What are the chemical elements in iced tea?
Natali5045456 [20]

Explanation:

I hope you interested about the chemical what they add in tea

6 0
2 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
2. Calculate the joules of energy required to heat 454.3g of water<br> frum 5.4°C to 98.6°C.
Tasya [4]

Answer:

176984.38J

Explanation:

E = mC∆T

Where E is the energy in joules

M is the mass of water

C is the specific heat capacity of water =4.184J/g°C

It is known that it will take 4.184J of energy to change the temperature of water by one degree Celsius.

∆T = 98.6°c - 5.4°c

= 93.2°c

∆H = 454.3g × 4.18J/g°C × 93.2°c

= 176984.3768

176984.38J

7 0
3 years ago
If I had a mixture of salt, sand, and iron filings I could separate the mixture by …
liubo4ka [24]
Okay to separate this mixture you will have to first use a magnet because the magnet will attract the iron filings out of the sand then you will remain with the sand and salt.

Next step is to add water to the sand and salt mixture the salt will dissolve in the water and the sand will not forming a heterogeneous mixture.Then using the method of filtration you can separate the sand from  the salt solution leaving the salt solution

 To separate the salt from the water use the method of evaporation this will enable you to get back the salt 

In the end you should beable to have the sand,salt and iron filings separate.

Hope this was helpful to you.

Happy New Years !!! May your year be prosperous 






3 0
2 years ago
Which two conditions would result in the weakest electric force between
Sergeu [11.5K]

Answer:

object are conductor with unbalanced charges

8 0
3 years ago
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