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kykrilka [37]
2 years ago
8

How many moles are present in 450 grams of sodium phosphide, Na3PNa3P?

Chemistry
1 answer:
BaLLatris [955]2 years ago
5 0

Answer:

4.5moles

Explanation:

Given parameters:

Mass of Na₃P = ?

Unknown:

Number of moles  = ?

Solution;

The number of moles in a given substance can be derived from the expression below;

    Number of moles = \frac{mass}{molar mass}  

Molar mass of  Na₃P   = 3(23) + 31 = 100g/mol

   Number of moles  = \frac{450}{100}   = 4.5moles

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They start with  the numbers u need to know in order to slove the problem and there has to be a story behind it 
7 0
3 years ago
How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0
3 years ago
How many moles of xenon gas, Xe, would occopy 37.8L at stp
Charra [1.4K]

Answer:

The number of moles of  xenon are 1.69 mol.

Explanation:

Given data:

Number of moles of xenon = ?

Volume of gas = 37.8 L

Temperature = 273 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values in formula.

1 atm × 37.8 L = n ×  0.0821 atm.L/ mol.K   ×273 K

37.8 atm.L =  n × 22.413 atm.L/ mol.

n = 37.8 atm.L /  22.413 atm.L/ mol.

n = 1.69 mol

The number of moles of  xenon are 1.69.

8 0
3 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
Is salt a iconically bounded compound?
zvonat [6]

Answer:

i believe u mean "ionically". Yes it is.

Explanation:

3 0
2 years ago
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