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meriva
1 year ago
12

A 70.-kg person exposed to ⁹⁰Sr absorbs 6.0X10⁵ β⁻ particles, each with an energy of 8.74X10⁻¹⁴ J.(c) What is the equivalent dos

e in sieverts (Sv)?
Chemistry
1 answer:
postnew [5]1 year ago
7 0

A 70.-kg person exposed to ⁹⁰Sr absorbs 6.0X10⁵ β⁻ particles, each with an energy of 8.74X10⁻¹⁴ J.

<h3>What is β⁻ particles ?</h3>

A beta particle, also known as a beta ray or beta radiation (symbol ), is a highly energetic, swiftly moving electron or positron that is released during the radioactive disintegration of an atomic nucleus. Beta decay occurs in two ways: decay and + decay, which result in the production of electrons and positrons, respectively.

In air, beta particles with an energy of 0.5 MeV have a range of roughly one meter; the range is energy-dependent.

Ionizing radiation of the sort known as beta particles is regarded, for the purposes of radiation protection, as being more ionizing than gamma rays but less ionizing than alpha particles. The damage to live tissue increases as the ionizing effect increases, but so does the radiation's penetration power.

To learn more about β⁻ particles from the given link:

brainly.com/question/10111545

#SPJ4

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For a system, H2(g) + I2(g) ⇌ 2 HI(g), Kc = 62.9 at 750 K. 2.80 moles of HI were placed in a 10.0-liter container, brought up to
aev [14]

Answer:

d. [HI] > [H2]

Explanation:

The explanation at equilibrium is shown below:-

Data provided           H_2(g) + I(g) \rightleftharpoons 2HI_(g)

Initial concentration    -           -           \frac{2.80}{10} = 0.280 M

At equilibrium             x          x       0.280 - 2x

K_c = \frac{(HI)^2}{(H_2)(I_2)}  = 62.9

= \frac{(0.280 - 2x)^2}{x^2} = 62.9\\\\4x^2 - 1.12x + 0.0784 = 62.9x^2

After solve the above equation we will get

x = 0.0282 M

Therefore at equilibrium

[H_2] = [I_2] = x = 0.0282M\\\\

[HI] = 0.280 - 2x = 0.2236 M

Hence, the correct option is d.

7 0
3 years ago
A 74.28-g sample of ba(oh)2 is dissolved in enough water to make 2.450 liters of solution. how many ml of this solution must be
Marianna [84]

First let us calculate the initial molarity of the 2.45 L of solution. Molar mass = 171.34 g/mol

<span>moles Ba(OH)2 = 74.28 g * (1 mole / 171.34 g) =  0.4335 moles</span>

Molarity (M1) = 0.4335 moles / 2.45 L = 0.177 M

 

Now using the formula M1V1 = M2V2, we can calculate how much to dilute (V1):

0.177 * V1 = 0.1 * 1

V1 = 0.56 L

 

<span>Therefore 0.56 L of the initial solution must be diluted to 1 L to make 0.1 M</span>

5 0
3 years ago
Which terms are used to identify pure substances
Semenov [28]

Answer:

Color, odor, density, melting temperature, boiling temperature, and solubility are examples of physical properties. Physical properties can be used to identify a pure substance.

Explanation:

please mark brainliest! :)

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3 0
3 years ago
Read 2 more answers
How does increasing the temperature of a liquid below the boiling point affect
Vilka [71]
When the particles of a substance (usually a liquid) is heated up, its particles absorb the energy provided thereby increasing their kinetic energy resulting to more movement of the individual particles.
4 0
3 years ago
A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h
NikAS [45]

Nitrogen combine with hydrogen to produce ammonia \text{NH}_3 at a 1:3:2 ratio:

\text{N}_2 \; (g) + 3 \;  \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. 3 \; \text{mol} of hydrogen gas would have been consumed while 2 \; \text{mol} of ammonia would have been produced. The final mixture would therefore contain

  • 17 \; \text{mol} of \text{H}_2 \; (g) and
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Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

  • \begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.926 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.723 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  2.037 \times 10^{2} \; \text{kPa} \end{array}
6 0
3 years ago
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