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inysia [295]
3 years ago
10

A 5.00 gram sample of magnesium sulfate hydrate (epsom salt) is heated in the lab to form anhydrous magnesium sulfate. After hea

ting the substance, it is recorded that 2.86 grams of anhydrous magnesium sulfate remain. What is the chemical formula for this sample of magnesium sulfate hydrate
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

The answer to your question is  MgSO₄ 5H₂O

Explanation:

Data

mass of MgSO₄ = 2.86 g

mass of H₂O = 2.14 g (5 - 2.86)

Process

1.- Calculate the molecular mass of the compounds

MgSO₄ = 24 + 32 + (16 x 4) = 120

H₂O = 16 + 2 = 18

2.- Convert the grams obtain to moles

                        120 g of MgSO₄  --------------- 1 mol

                         2.8 g                  ----------------  x

                         x = (2.8 x 1)/120

                        x = 0.024 moles

                         18 g of H₂O --------------------- 1 mol

                         2.14 g           -------------------- x

                         x = (2.14 x 1)/18

                         x = 0.119

3.- Divide by the lowest number of moles

MgSO₄  = 0.024/0.024 = 1

H₂O = 0.119/ 0.024 = 5

4.- Write the molecular formula

                   MgSO₄5H₂O                        

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When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)
Firdavs [7]

Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ =  0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂  + 2K₃PO₄  → Ba₃(PO₄)₂  + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

              Ba(NO₃)₂        :         Ba₃(PO₄)₂

                   3                :               1

              0.182 × 10⁻³    :              1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

                K₃PO₄           :          Ba₃(PO₄)₂

                   2                 :                1

              2.537 × 10⁻³     :               1/2 ×  2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by  Ba(NO₃)₂  are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

6 0
3 years ago
What bonds are broken and octane combustion?
dedylja [7]

Answer:

Common combustion reactions break the bonds of hydrocarbon molecules,

Explanation:

the resulting water and carbon dioxide bonds always release more energy than was used to break the original hydrocarbon bonds. That's why burning materials mainly made up of hydrocarbons produces energy and is exothermic.

7 0
2 years ago
What types of properties change during a chemical reaction?
ryzh [129]
The chemical composition 

4 0
3 years ago
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In which of the following phases of matter do molecules have the lowest amount of energy?
Elden [556K]

Answer:

C. Solid

Explanation:

It's most about the kinetic energy, when molecules have the least amount of room to move around, they have the least amount of energy.

For example if you think about gas and how spread out the molecules are in order to evaporate, there it the highest amount of energy there.

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3 years ago
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How many milliliters of 0.50 M KOH are needed
spayn [35]

Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

Molarity of base, KOH (Mb) = 0.5M

Volume of acid, HNO3 (Va) = 10mL

Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

1.5 x 10 / 0.5 x Vb = 1

Cross multiply

0.5 x Vb = 1.5 x 10

Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

3 0
3 years ago
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