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Setler79 [48]
3 years ago
9

How many carbon monoxide molecules are in 0.75 moles of carbon monoxide

Chemistry
2 answers:
Naddika [18.5K]3 years ago
4 0

Hello!

Firstly, we need to determine the atomic formula of carbon monoxide.

When breaking down the name, we know that there is only one carbon atom since we normally do not add the prefix mono- (one), in the front of the first element. Secondly, there is one oxygen atom since mono = one, and oxide is oxygen. Carbon monoxide is written as: CO.

Now, to find the number of molecules, we will need to use a conversion factor, and also we must know Avogadro's number: <u>6.02 x 10^23 molecules</u>.

To set up the conversion factor, we must change the units from "moles CO" to "molecules CO". To do that, we will begin with the given number, "0.75 moles CO", and multiply that by "6.02 x 10^23 molecules CO over 1 mol CO".

0.75 moles CO x (6.02 x 10^23 molecules CO / 1 mol CO) = 4.52 x 10^23 molecules CO

Therefore, there are about 4.52 x 10^23 molecules CO.

masya89 [10]3 years ago
3 0
0.75 x 6.02x 10^23
No of molecules = no of moles x avogadro’s number
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As here, we can use the pK_{a} nearest to the desired pH.

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

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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

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

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

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What mass of water could be warmed from 21.4 degrees celsius to 43.4 degrees celsius by the pellet dropped inside it? Heat capac
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42.34 g of water could be warmed from 21.4°C to 43.4°C  by the pellet dropped inside it

Heat loss by the pellet is equal to the Heat gained by the water.

q_{w} = -q_{p} ….(1)

where, q_{w} is the heat gained by water

q_{p} is the heat loss by pellet

q_{w} = mCΔT

where m = mass of water

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

ΔT = Increase in temperature

ΔT for water = 43.4 - 21.4 = 22°C

q_{w} = m × 4.184 × 22 …. (2)

Now

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Learn more about specific heat here brainly.com/question/16559442

#SPJ1

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<h3>Heat Transfer</h3>

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\delta T = (400 - 273) - (263 - 273) = \\&#10;\delta T = T_2 - T_1\\&#10;\delta T = -10 - 127\\&#10;\delta T = -137^0C

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Learn more on heat transfer here;

brainly.com/question/16055406

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