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nata0808 [166]
1 year ago
14

an unknown amount of mercury (ii) oxide was decomposed in the lab. mercury metal was formed and 4.50 l of oxygen gas was release

d at a pressure of 0.970 atm and 390.0 k. what was the initial weight of mercury oxide in the sample?
Chemistry
1 answer:
Assoli18 [71]1 year ago
4 0

The initial weight of mercury oxide in the sample was 59.1 g HgO.

<h3>Steps</h3>

chemical reaction

2HgO ⟶ 2Hg + O₂

the moles of O₂

pV = nRT

n = (pV)/(RT)

Data:

p = 0.970 atm

V = 4.50 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 390.0 K

Calculation:

n = (0.970 × 4.500)/(0.082 06 × 390.0)

n = 0.1364 mol O₂

the moles of HgO

The molar ratio is 1 mol O₂/2 mol HgO.

Moles of HgO = 0.1364 mol O₂ × (2 mol Hg/1 mol O₂)

Moles of HgO = 0.2728 mol HgO

the mass of HgO

Mass of HgO = 0.2728 mol HgO × (216.59 g HgO/1 mol HgO)

Mass of HgO = 59.1 g HgO

<h3>What is the name of HgO?</h3>

For the creation of various organic mercury compounds and specific inorganic mercury salts, mercury(II) oxide, or HgO, serves as a source of elemental mercury.

This red or yellow crystalline substance is also utilised in mercury batteries and zinc-mercuric oxide electric cells as an electrode (combined with graphite).

<h3>What is the purpose of mercury oxide?</h3>

Mercuric oxide is a colourless, crystalline powder that ranges from yellow to orange-yellow.

It serves as a seed protectant, a pigment, a preservative, and an ingredient in alkaline batteries and cosmetics.

<h3>Is there a combination of mercury oxide?</h3>
  • The powder form of mercury oxide is dark black or dark brown.
  • An intimate blend of metallic mercury and mercuric oxide rather than a genuine compound.

learn more about mercury oxide here

brainly.com/question/3235037

#SPJ4

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What is the volume of an irregular object if the initial volume in the graduated cylinder is 3.50 ml and it rises to 7.50 ml aft
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Answer is: the volume of an irregular object is 4,00 ml.
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</span>V(irregular object) = V(final volume) - V(initial volume).
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8 0
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Think about the lab procedure you just read. Label each factor below with V for “variable” or C for “constant”.
luda_lava [24]
<h2>Answer:</h2><h3>The temperature of the gas: V</h3>

The temperature of gas is a variable quantity. It can be changed by changing energy or pressure of gas.

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It is a constant entity. As mass of gas once taken can not be changed by changing temperature, pressure etc.

<h3>The radius of the tube: C</h3>

The radius of tube cannot change at any rate.

<h3>The temperature of the gas (changed by the water surrounding it):  V</h3>

It can be changed by changing the temperature of water surrounding it.

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Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
avanturin [10]
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
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