Answer is: 1. HClO₃; (chloric acid).
Balance chemical reaction (dissociation):
ClO₃⁻(aq) + H₂O(l) ⇄ HClO₃(aq) + OH⁻(aq).
According
to Bronsted-Lowry theory acid are donor of protons and bases
are acceptors of protons (the hydrogen cation or H⁺).
The chlorate anion (ClO₃⁻) is Bronsted base and it
can accept proton and become conjugate acid HClO₃..
Answer:
15 grams of water
Explanation:
15 grams of water of water would lose heat the faster compared to higher masses of water.
Water generally is a poor conductor heat.
- To heat up a unit of water, significant amount of energy must be added to the body of water.
- With time, the body continues to increase in temperature.
- A 500g mass of water will take more time to lose heat.
Answer:
Explanation:
Bronsted Base is an H+ acceptor
No good answer Bronstead base does not accept hydroxide or electrons
Answer:
A. Energy is transferred to different forms
.
Explanation:
Hello!
In this case, we need to consider the law of conservation of mass and energy which states that mass and energy cannot be neither created nor destroyed, just modified; it means we can rule out B. and C. so far.
Moreover, since D. is actually true for combustion reactions because they are used to provide energy in industrial operations, this is not the concern here because a combustion reaction is not considered.
Therefore the correct option is A. Energy is transferred to different forms as the energy provided by Rose is transferred to the pendulum system
.
Best regards!
Answer:
vHe / vNe = 2.24
Explanation:
To obtain the velocity of an ideal gas you must use the formula:
v = √3RT / √M
Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:
vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol
vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
The ratio is:
vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol
<em>vHe / vNe = 2.24</em>
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I hope it helps!