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erica [24]
3 years ago
5

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.90 moles of magnesium perchlorate, Mg(ClO4)2.

Chemistry
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

5.90, 11.8, 47.2

Explanation:

Let’s remove the parentheses and write the formula as MgCl₂O₈.

We see that 1 mol Mg(ClO₄)₂ contains 1 mol Mg atoms, 2 mol Cl atoms, and 8 mol O atoms.

∴ \text{Moles of Mg atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{1 mol Mg atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{5.90 mol Mg atoms}  

\text{Moles of Cl atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{2 mol Cl atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{11.8 mol Cl atoms}

\text{Moles of O atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{8 mol O atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{47.2 mol O atoms}

∴ Mg, Cl, O = 5.90, 11.8, 47.2

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Answer:

the solvent is water and the solute is salt

Explanation:

A solute is the component in a solution in the lesser amount. In a NaCl solution, the salt is the solute. A solution may contain more than one solute.

7 0
3 years ago
A gas expands and does P-V work on the surroundings equal to 275 J. At the same time, it absorbs 216 J of heat from the surround
mafiozo [28]

Answer:

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Explanation:

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7 0
2 years ago
Which of the following contains a covalent bond?<br> A) BeO<br> B) LiCI<br> C) CN<br> D) MgO
Svetllana [295]

Answer= C) CN contains a covalent bond.

7 0
2 years ago
22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass
Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃  +  MgBr₂  → Mg(NO₃)₂  +  2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0
2 years ago
In step 2, of the experiment, the procedure uses 3.0M NaOH. However, the student notices that the only solution of NaOH is conce
Luda [366]

Answer:

We need 78.9 mL of the 19.0 M NaOH solution

Explanation:

Step 1: Data given

Molarity of the original NaOH solution = 19.0 M

Molarity of the NaOH solution we want to prepare = 3.0 M

Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

Step 2: Calculate volume of the 19.0 M NaOH solution needed

C1*V1 = C2*V2

⇒with C1 = the concentration of the original NaOH solution = 19.0 M

⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED

⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M

⇒with V2 = the volume  of the NaOH solution we want to prepare = 500 mL = 0.500 L

19.0 M * V2 = 3.0 M * 0.500 L

V2 = (3.0 M * 0.500L) / 19.0 M

V2 = 0.0789 L

We need 0.0789 L

This is 0.0789 * 10^3 mL = 78.9 mL

We need 78.9 mL of the 19.0 M NaOH solution

8 0
3 years ago
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