Answer:
0.3267 M
Explanation:
To solve this problem, first we calculate how many moles of Mn(ClO₄)₂ are contained in 23.640 g of Mn(ClO₄)₂·6H₂O.
Keep in mind that the crystals of Mn(ClO₄)₂ are hydrated, and <em>we need to consider those six water molecules when calculating the molar mass of the crystals</em>.
Molar mass of Mn(ClO₄)₂·6H₂O = 54.94 + (35.45+16*4)*2 + 6*18 = 361.84 g/mol
Now we <u>proceed to calculate</u>:
- 23.640 g Mn(ClO₄)₂·6H₂O ÷ 361.84 g/mol = 0.0653 mol Mn(ClO₄)₂·6H₂O = mol Mn(ClO₄)₂
Now we divide the moles by the volume, to <u>calculate molarity</u>:
- 200 mL⇒ 200/1000 = 0.200 L
- 0.0653 mol Mn(ClO₄)₂ / 0.200 L = 0.3267 M
Answer:
Atom is a the smallest particle of a chemical element that can exist.
Molecules is a a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
Particles is a minute portion of matter.
A particle can be a single atom or a molecule ( a group of atoms held together by chemical bonds).
Answer:
See explanation.
Explanation:
Hello there!
In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Thus, we proceed as follows:
Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

Best regards!
The closeness of a measurement to its true value is a measure of its Accuracy.
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>