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3 years ago

If volume is increased, the number of collisions per second

Chemistry

1 answer:

frutty [35]3 years ago

5 0

Answer:

Decreases

Explanation:

The particles have more space to move and will be less likely to collide.

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Can you tell me the simple explanation what the mol in chemistry is?

Setler [38]

By the way it's not mol, it's mole.
The mole is the unit of amount in chemistry. It provides a bridge between the atom and the macroscopic amounts of material that we work with in the laboratory. It allows the chemist to weigh out amounts of two substances, say iron and sulfur, such that equal numbers of atoms of iron and sulfur are obtained. A mole of a substance is defined as:The mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of 12C.

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3 years ago

Which of these is a benefit of nuclear fusion?

dybincka [34]

Answer:

B. There is a lower risk of runaway chain reactions.

Explanation:

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2 years ago

How does the polarity of water give it this solvent property

mafiozo [28]

Answer:

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3 years ago

Read 2 more answers

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

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3 years ago

Describe what happens when a bond is created between magnesium and bromine. Be specific and explain in terms of electrons.

Allushta [10]

Answer: Magnesium and Bromine/MgBr2 = Ionic compounds

Explanation: When atoms form together they can form between Ionic Compounds and molecules; this could depend on if they're joined by Covalent bonds as well because when atoms form with Covalent bonds, it forms Molecules.

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3 years ago