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alukav5142 [94]
1 year ago
9

The isotope \left.\begin{array}{r}212 \\ 83\end{array}\right? Bi has a half-life of 1.01 yr. What mass (in mg) of a 2.00-mg samp

le will remain after 3.75X10³ h?
Chemistry
1 answer:
anyanavicka [17]1 year ago
5 0

Half-life is the length of time it takes for half of the radioactive atoms of a specific radionuclide to decay. A good rule of thumb is that, after seven half-lives, you will have less than one percent of the original amount of radiation.

<h3>What do you mean by half-life?</h3>

half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive.

<h3>What affects the half-life of an isotope?</h3>

Since the chemical bonding between atoms involves the deformation of atomic electron wavefunctions, the radioactive half-life of an atom can depend on how it is bonded to other atoms. Simply by changing the neighboring atoms that are bonded to a radioactive isotope, we can change its half-life.

Learn more about half life of an isotope here:

<h3>brainly.com/question/13979590</h3><h3 /><h3>#SPJ4</h3>
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What is the cause of acid precipitation?
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Answer:

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Explanation:

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What data do you need to determine the specific heat capacity of a substance?
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Speific heat capacity is measured with the aid of determining how a whole lot warmth electricity is needed to increase one gram of a substance one digree Celsius. The Speific heat capacity of water is 4.2 joules per gram per degree Celsius or 1 calorie in step with gram per digree Celsius.

The specific heat capacity is defined as the amount of heat (J) absorbed consistent with unit mass (kg) of the substance while its temperature increases 1 ok (or 1 °C), and its units are J/(kg k) or J/(kg °C).

Factors specific heat capacity relate to are temperature and strength.

The Speific heat capacity C can be measured as q = mC∆T

Or, C = q/m∆T

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∆T is the change in temperature

As a consequence so as to degree the specific heat capacity we need to recognize mass of the substance, quantity of heat lost or gain by the substance and the exchange in temperature.

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6 0
2 years ago
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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3 years ago
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jekas [21]
D is the correct answer... if u need in depth let me know
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