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mart [117]
3 years ago
14

Select the two variables that are held constant when testing Boyle's law in a manometer.

Chemistry
2 answers:
MrRissso [65]3 years ago
4 0
Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select<span> all that apply. V2 = k/P2 V2 = P1V1/P2 ... What </span>two variables<span> are </span>held constant when testing Boyle's Law in a manometer<span>? Temperature hope this helps

</span>
gladu [14]3 years ago
4 0

<u>Answer:</u> The variables which remain constant while testing Boyle's law is temperature and number of moles.

<u>Explanation:</u>

Boyle's law states that pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

Mathematically,

P\propto \frac{1}{V}     (At constant temperature and number of moles)

or,

PV=k

With increase in pressure, the volume of the gas decreases and vice-versa.

Hence, the variables which remain constant while testing Boyle's law is temperature and number of moles.

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Why does heating increase the speed at which a solute dissolves in water?
saw5 [17]

Answer: C

Explanation:

It gives kinetic energy to the molecules and it breaks the bonds faster because they jiggle more

8 0
3 years ago
Briefly explain why an unbalanced chemical equation cannot fully describe a reaction?
zubka84 [21]

From the conservation of mass, matter is not created nor destroyed. So ponder it this way, for example I put one piece of toast in a toaster, but when it is done I had two pieces, that makes no sense right? Same thing with a chemical reaction but in molecular terms, all things must stay constant. If you put a convinced amount you have to get that quantity back, not less or not more in broad terms.

5 0
3 years ago
How did the concentration of iodide ions change in these two trials, how did the rate change accordingly?
AfilCa [17]
<span>The concentration of iodide ions doubled.
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3 0
3 years ago
2. A solution is 0.01 M in Ba2+ and 0.01 M in Ca2+. Sodium sulfate is added to selectively precipitate one of the cations, while
lozanna [386]

Answer:

Explanation:

The solution contain  0.01 M concentration of Ba²⁺

0.01M concentration of Ca²⁺

Ksp ( solubility constant) for BaSO₄ = 1.07 × 10⁻¹⁰

Ksp for CaSO₄ = 7.10 × 10⁻⁵

(BaSO₄) = (Ba²⁺) (SO₄²⁻)

1.07 × 10⁻¹⁰ = 0.01 M (SO₄²⁻)

1.07 × 10⁻¹⁰ / 0.01 = ( SO₄²⁻)

1.07 × 10⁻⁸ M = ( SO₄²⁻)

so the minimum of concentration of concentration sulfate needed is 1.07 × 10⁻⁸ M

For CaSO₄

CaSO₄ = ( Ca²⁺) (  SO₄²⁻)

7.10 × 10⁻⁵  = 0.01 (SO₄²⁻)

(SO₄²⁻) = 7.10 × 10⁻⁵   / 0.01 = 7.10 × 10⁻³ M

so BaSO₄ will precipitate first since its cation (0.01 M Ba²⁺) required a less concentration of SO₄²⁻ (1.07 × 10⁻⁸ M ) compared to CaSO₄

b) The minimum concentration of SO₄²⁻ that will trigger the precipitation of the cation ( 0.01 M Ba²⁺) that precipitates first is 1.07 × 10⁻⁸ M

7 0
4 years ago
A 6.13 g sample of an unknown salt (MM = 116.82
Olin [163]

Answer:

-3.19x10³ J

Explanation:

Since the surroundings absorbed 3.19 × 10³ J (or 3190 J) of heat, the system, or the dissolution reaction, must have lost the same amount of heat. The heat for the system, then, is -3.19 × 10³ J (or -3190 J). We know this is true because of the first law of thermodynamics, "heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy".

6 0
3 years ago
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