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1 month ago

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What phenomenon in nature do scientists use to define the length of a meter? Why is this a better definition than “the length of

a standard meter stick”?

1 answer:

eduard [2.6K]1 month ago

8 0

Answer:

The speed of light serves as the natural phenomenon that defines the meter's length. A meter is quantified as the distance light travels in a vacuum in 1/(299792458) seconds.

This definition is superior because using a standard meter stick can lead to inaccuracies due to variations in measurements resulting from different atmospheric conditions.

Explanation:

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"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

lorasvet [2668]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

From the information provided:

Let the fractional abundance for $_{14}^{28}\textrm{Si}$ isotope be 'x'

For $_{14}^{28}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 27.9769 amu

Percentage abundance of $_{14}^{28}\textrm{Si}$ isotope = 92.22 %

Fractional abundance for $_{14}^{28}\textrm{Si}$ isotope = 0.9222

For $_{14}^{29}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 28.9764 amu

Percentage abundance for $_{14}^{28}\textrm{Si}$ isotope = 4.68%

Fractional abundance of $_{14}^{28}\textrm{Si}$ isotope = 0.0468

For $_{14}^{30}\textrm{Si}$ isotope:

Mass of $_{14}^{30}\textrm{Si}$ isotope = 29.9737 amu

Fractional abundance for $_{14}^{30}\textrm{Si}$ isotope = x

The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

$28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309$

To convert this fractional abundance into a percentage, multiply by 100:

$\Rightarrow 0.0309\times 100=3.09\%$

This shows that the percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

6 0

1 month ago

In how many grams of water should 25.31 g of potassium nitrate (kno3) be dissolved to prepare a 0.1982 m solution?

lions [2782]

Solution:

Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.

Thus,

Molality = moles of solute / kg of solvent.

Therefore, kg of solvent = moles of solute / molality.

moles of solute = mass / molar mass

= 25.31 g / 101.1 g/mole

= 0.2503 mole.

kg of solvent = 0.2503 mole / 0.1982 m

= 1.263 kg

= 1263 g.

This is the final answer.

6 0

1 month ago

Read 2 more answers

The value of delta G at 141.0 degrees celsius for the formation of phosphorous trichloride from its constituent elements,

VMariaS [2860]

The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.

6 0

14 days ago

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

eduard [2645]

Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

Rationale:

Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

8 0

1 month ago

In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a

KiRa [2853]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

$\Delta H =\frac{Q}{n}$

To calculate the heat (Q):

$Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J$

Next, we need to calculate the number of moles involved in the reaction (n):

$n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles$

With these two values, we can substitute them into the first equation:

$\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole$

Have a great day!

5 0

1 month ago