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Murrr4er [49]
3 years ago
7

What is the mass of 1 mole of MgSO4?

Chemistry
1 answer:
White raven [17]3 years ago
7 0

Answer:

120.37 g

Explanation:

Mg = 24.31 g

S = 32.06 g

4 O = 16 X 4 = 64 g

Therefore, MgSO4 is (24.31 + 32.06 + 64) g = 120.37 g

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Passes through the kidneys and the good stuff is kept, and the other stuff is passed through leading to the bladder.  The once the bladder is full enough to send the message to the brain, urine (waste) is eliminated from the body.
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How many total atoms are in 0.290 g of P2O5?
Sliva [168]

There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\

= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\

Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

∴ \text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\

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3 years ago
During a laboratory experiment, 36.12 grams of Aboz was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What
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Answer:

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First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:

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Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:

  • 0.354 mol AI₂O₃ * \frac{3molO_2}{2molAl_2O_3} = 0.531 mol O₂

We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:

  • 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
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