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2 years ago

You need 2.5 moles of aluminum for an experiment. How many atoms of aluminum is this?

Chemistry

1 answer:

emmainna [20.7K]2 years ago

3 0

Answer:

1.505×10^24 atoms

Explanation:

1mole=6.02×10^23 atoms

2.5moles=?

(crossmultiply)

2.5×6.02×10^23 ÷1

=1.505×10^24 atoms

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3 years ago

1. What did Henri Becquerel discover about radiation emitted from uranium salts?

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2 years ago

You were instructed to add 1.0 mL out of 4.0 mL of an undiluted sample to 99 mL of sterile diluent. Instead, you add the entire

finlep [7]

Answer:

A.) Intended dilution factor : 0.01

B.) Actual dilution factor: 0.039

Explanation:

Dilution factor(DF) is ratio of the volume of the undiluted sample to total volume of diluted solution

A.) What was your intended dilution factor

This would be the the addition 1.0ml from 4.0ml to 99ml of sterile

Volume of sterile diluent = 99ml

Volume of Undiluted sample = 1.0ml

Total volume of diluted solution = 99ml + 1m = 100ml

Dilution factor = Volume of undiluted sample ÷ Total volume of diluted solution

Dilution factor = 1:100

= 1ml ÷ 100ml

= 0.01

B.) What was your actual dilution factor?

This would be the entire addition of 4.0ml to 99ml sterile diluent

Volume of sterile diluent = 99ml

Volume of Undiluted sample = 4.0ml

Total volume of diluted solution = 99ml + 4.0ml = 103ml

Dilution factor = Volume of undiluted sample ÷ Total volume of diluted solution

Dilution factor

= 4:103

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= 0.039

8 0

2 years ago

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 $1.87\times 10^{-14}M$

7 0

2 years ago

You need 2.5 moles of aluminum for an experiment. How many atoms of aluminum is this?​

You need 2.5 moles of aluminum for an experiment. How many atoms of aluminum is this?