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Viefleur [7K]
3 years ago
12

Food and medical procedures can expose people to radiation. Therefore, the effects of radiation from a nuclear power plant are n

ot a reason for concern.
Chemistry
2 answers:
iogann1982 [59]3 years ago
7 0
False, the radiation from food and medical procedures have very little consequences. The radiation from a nuclear power plant can kill people which makes it a very large problem if not contained properly.
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Mariana [72]3 years ago
5 0

Answer:

It's false

Explanation:

The radioactive materials of a nuclear plant emit ionizing radiation that can damage tissues. The radiation dose of a person is approximately 2.5 millisieverts per year by radiation from natural sources. Workers in a nuclear industry are exposed to approximately 4.5 millisieverts. Ionizing radiation transfers energy to cell molecules in tissues. As a consequence, these cell functions can be damaged temporarily or permanently. The severity depends on the type of radiation, the dose and the rate of absorption.

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If you have 15g of sodium carbonate, what<br> mass of sodium is present in that sample?
Sati [7]

Answer: 6.75 g

Explanation:

5 0
3 years ago
PLEASE HELP 30 POINTS PLS PLS
joja [24]
22-yes, they belong to the same group
23- they are metals
24-Atoms of group 1 elements all have one electron in their outer shell. This means Na and K have the same valence electrons.
25-Metals are solids at room temperature, therefore Na and K are solid.
26- mass number=A=235
Protons=z=92
Neutrons=mass number-protons
Neutrons=A-Z
Neutrons=235-92
Neutrons=143
8 0
3 years ago
How many grams of NaHCO3 are needed to prepare 250 mL of 0.50 M<br> NaHCO3?
forsale [732]

Answer:

10.5g

Explanation:

First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:

Volume = 250mL = 250/1000 = 0.25L

Molarity = 0.5M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.5 x 0.25

Mole = 0.125 mole

Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:

Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol

Number of mole of NaHCO3 = 0.125 mole

Mass of NaHCO3 =?

Mass = number of mole x molar Mass

Mass of NaHCO3 = 0.125 x 84

Mass of NaHCO3 = 10.5g

Therefore, 10.5g of NaHCO3 is needed.

6 0
2 years ago
I NEED THIS ASAP!!! Why is it important to know if a substance produced by chemical industry is pure?
ratelena [41]

Answer:High-purity chemicals are all-the-more important during the manufacturing process because chemicals are often used in bulk. ... In plastics, for example, an impure chemical substrate can produce an end-product that's too brittle to use. In pharmaceuticals, one impure component can render a drug dangerous

Explanation:

6 0
3 years ago
Write packing efficiency of fcc ,BCC, SCC and formula also . ?​
Minchanka [31]

❖ <u>Packing Efficiency</u> ❖

➪ The percentage of total space occupied by particles is called <u>packing efficiency</u>.

\\ \qquad{\rule{200pt}{3pt}}

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

❖ Packing efficiency of simple cubic structure (SCC).❖

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

:  : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3

  • Volume of unit cell = a³

  • Volume of unit cell = (2r)³

  • Volume of unit cell = 8r³

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪ Substituting the known values in the formula, we get the following results:

:  : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ :  : \implies   \dfrac{\pi}{6} \times 100 \\\\\   :  : \implies \bf {52.4 \%}

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

\\ \qquad{\rule{200pt}{3pt}}

➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3

  • Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪Substituting the known values in the formula, we get the following results:

:   : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\  \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\  \implies\bf52.4 \%

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

\\ \qquad{\rule{200pt}{3pt}}

❖ Packing efficiency of body cubic structure (BCC).

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

: : \implies  \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3

  • Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪Substituting the known values in the formula, we get the following results:

: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\  \implies \dfrac{3\pi}{8} \times 100 \\\ \\  \bf  \implies \: 68 \%

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

\\ \qquad{\rule{200pt}{3pt}}

6 0
2 years ago
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