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svp [43]
3 years ago
6

AV 5. How many grams are in 0.52 moles of H3PO4?

Chemistry
1 answer:
Hoochie [10]3 years ago
3 0

Answer:

50.96g

Explanation:

Given parameters:

Number of moles of H₃PO₄   = 0.52moles

Unknown:

Mass of the compound  = ?

Solution:

To find the mass of the compound:

    Mass  = number of moles x molar mass of H₃PO₄

Molar mass of H₃PO₄ = 3(1) + 31 + 4(16)  = 98g/mol

  Mass  = 0.52 x 98  = 50.96g

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Vikki [24]
34.95 atm

lol i hope i’m not too late



4 0
3 years ago
Write the molecular formula for the following compound.<br>​
Sholpan [36]

Answer : The molecular formula of the compound is, C_8H_9O_3N

Explanation :

Molecular formula : It is the representation of substance by the symbols and it denotes the number of atoms of each element present in the compound.

Now count the number of carbon, hydrogen, nitrogen and oxygen atoms present in the given compound.

As we see that in the given compound, there are 8 atoms of carbon element, 3 atoms of oxygen element, 1 atom of nitrogen element, 9 atoms of hydrogen element.

Thus, the molecular formula of the compound will be C_8H_9O_3N

7 0
3 years ago
The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of CO2 genera
nevsk [136]

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate =3.45 g\times \frac{32.5}{100}=1.121 mol

Moles of bicarbonate ion = \frac{1.121 g/mol}{61 g/mol}=0.01840 mol

HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

\frac{1}{1}\times 0.01840 mol=0.01840 mol of carbon dioxide gas

Moles of carbon dioxide gas  n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

PV=nRT (ideal gas equation)

V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

5 0
3 years ago
PLEASE HELP WITH HIS DUE SOON WILL GIVE BRAINLIST HELP!!
Readme [11.4K]

Answer:

320

Explanation:

8*8*5

8 0
3 years ago
Which element would be a strong reducing agent?<br><br><br> F2<br><br> Ba<br><br> Na<br><br> Cl2
Damm [24]
F2 because it’s the strongest
3 0
3 years ago
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