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2 years ago

We can determine the potential of a single electrode by:____________.

Chemistry

1 answer:

postnew [5]2 years ago

6 0

Answer:

d. none of the above

Explanation:

An electrochemical cell consists of two half-cells, with each cell containing an electrode.

The potential of a single electrode in a half-cell is called the Single electrode potential.

The emf of a cell that consists of two half-cells can be determined by connecting them to a voltmeter.

However, there is no way of measuring the emf of a single half-cell directly.

The only way to determine the emf of a single electrode is to combine it with a standard hydrogen electrode (SHE) and measure it with a voltmeter.

Therefore, the correct option is D "none of the above"

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A scientist has two samples of gas: The first sample contains one mole of argon atoms and has a mass of 39.948 g; the second sam

V125BC [204]

There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.

Explanation:

39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.

4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.

We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.

So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.

Learn more about:

Avogadro's number

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3 years ago

If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $Al_2O_3$ = $\frac{16.9\times 1000g}{102g/mol}=165.7moles$

moles of $NaOH$ = $\frac{57.4\times 1000g}{40g/mol}=1435moles$

moles of $HF$ = $\frac{57.4\times 1000g}{20g/mol}=2870moles$

As 1 mole of $Al_2O_3$ reacts with 6 moles of $NaOH$

166 moles of $Al_2O_3$ reacts with = $\frac{6}{1}\times 166=996$ moles of $NaOH$

As 1 mole of $Al_2O_3$ reacts with 12 moles of $HF$

166 moles of $Al_2O_3$ reacts with = $\frac{12}{1}\times 166=1992$ moles of $HF$

Thus $Al_2O_3$ is the limiting reagent.

As 1 mole of $Al_2O_3$ produces = 2 moles of cryolite

166 moles of $Al_2O_3$ reacts with = $\frac{2}{1}\times 166=332$ moles of cryolite

Mass of cryolite $(Na_3AlF_6)$ = $moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg$

Thus 69.72 kg of cryolite will be produced.

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2 years ago

The radioactive isotope I-131 is used in

Fudgin [204]

The answer is D, medical diagnosis

3 0

2 years ago

How will an increase in wind speed affect soil erosion?

AnnyKZ [126]

Answer:

A. Soil erosion will increase.

7 0

2 years ago

A balloon containing helium gas expands from 230

Anit [1.1K]

The answer for the following problem is mentioned below.

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

Explanation:

Given:

Initial volume ( $V_{1}$ ) = 230 ml

Final volume ( $V_{2}$ ) = 860 ml

Initial moles ( $n_{1}$ ) = 3.8 × $10^{-4}$ moles

To find:

Final moles ( $n_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

V ∝ n

$\frac{V_{1} }{V_{2} }$ = $\frac{n_{1} }{n_{2} }$

where,

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $n_{1}$ ) represents the initial moles of the gas

( $n_{2}$ ) represents the final moles of the gas

Substituting the above values;

$\frac{230}{860}$ = $\frac{3.8 * 10^-4}{n_{2} }$

$n_{2}$ = 14.2 × $10^{-4}$ moles

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

7 0

2 years ago