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3 years ago

Which label belongs in the region marked X?

Chemistry

1 answer:

lubasha [3.4K]3 years ago

7 0

Answer:

A) involves changes in temperature

Explanation:

The figure is missing, but I assume that the region marked X represents the region in common between Gay-Lussac's law and Charle's Law.

Gay-Lussac's law states that:

"For an ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$p\propto T$

where p is the pressure of the gas and T its absolute temperature.

Charle's Law states that:

"For an ideal gas kept at constant pressure, the volume of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$V\propto T$

where V is the volume of the gas and T its absolute temperature.

By looking at the two descriptions of the law, we see immediately that the property that they have in common is

A) involves changes in temperature

Since the temperature is NOT kept constant in the two laws.

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When a piece of sodium carbonate reacts with dilute HCl, a gas X is produced. When gas X is passed through lime water then the s

siniylev [52]

Gas x would be carbon dioxide.

note/ acid + carbonate —> salt + water + carbon dioxide

the white precipitate would be calcium carbonate. CaCo₃

note/ this is a common eqn u need to remember.

X - CO₂ (carbón dioxide)
Y - CaCo₃ (calcium carbonate)

sodium carbonate is a basic salt

7 0

3 years ago

Density is the _____ of a certain ______

olga2289 [7]

I think the first blank is "solidity" and second is "object."

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3 years ago

What is the significance of equivalence point in acid base titrations?

iren2701 [21]

Hey there!

The equivalence is point in a titration is the point at which you have neutralized all of your base/acid with your titrant acid/base from a buret. This can be seen with indicators which change color at the equivalence point in a titration to signal to you that all of your base/acid has been reacted with. For example, all your molecules of OH⁻ from a NaOH base in a beaker have been neutralized by H⁺of HCl acid from your titrant in a buret leaving only Na⁺ ions and Cl⁻ ions and neutral H₂O molecules.

7 0

3 years ago

7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat

Pie

Answer:

11.4g of S₂Cl₂ is the expected yield

9.69g of S₂Cl₂ are produced with a 85% yield

Explanation:

The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:

S₈ + 4Cl₂ → 4S₂Cl₂

<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>

To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:

S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈

Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂

For a complete reaction of 0.0390 moles of sulfur, there are necessaries:

0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.

As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:

0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =

<h3>11.4g of S₂Cl₂ is the expected yield</h3>

If you produce just the 85.0% of yield, mass of S₂Cl₂ is:

11.4g ₓ 85% =

3 0

3 years ago

Find the oxidizing agent and the reducing agent.

Vladimir [108]

Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.

(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.

The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.

(2) C(s) + O2(g) --> CO2(g)

The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.

The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.

(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺

The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.

The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.

8 0

3 years ago