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3 years ago

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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Vlad [161]

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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3 years ago

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 grams of P₂O₅ to moles of P₂O₅.

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

Step 2. Convert moles of P₂O₅ to molecules of P₂O₅.

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

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

Step 3. Convert molecules of P₂O₅ to atoms.

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

kvv77 [185]

Answer:

8.70 liters

Explanation:

3O₂+ 4Al → 2AI₂O₃

First we convert 36.12 g of AI₂O₃ into moles, using its molar mass:

36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃

Then we convert AI₂O₃ moles into O₂ moles, 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 PV=nRT equation to calculate the volume, V:

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

V = 8.708 L

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2 years ago