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3 years ago

Which type of radioactive decay can pass through the body?

Chemistry

1 answer:

Mkey [24]3 years ago

3 0

Q. Which type of radioactive decay can pass through the body?

A. Gamma Rays

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Help me pls I put 49 point( every single point that I have) pls help me

Aliun [14]

Answer:

Chemical reaction, compound, molecule, covalent bonds, ionic bonds, ions. (first page)

Oxygen and hydrogen, 1 hydrogen 2 oxygen, H2O, a covalent bond, then mark off is always a liquid. (Second page)

Explanation:

6 0

2 years ago

What is an environmental hazard created by humans?

Yuliya22 [10]

Environmental Hazards are usually any chemicals that donot naturally occur to exist anywhere and are usually made in the fields of industry or experimental sciences
So if you happen for example to throw a bit of mercury in some river while being in a school trip then this is an environmental hazard created by humans

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3 years ago

Which of the following latitudes receive the most direct solar energy?

Misha Larkins [42]

Answer:

The answer to your question is D.

Explanation:

The latitudes near the equator receives the most direct solar energy.

Hope this helps :)

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

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3 years ago

For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the

Aleks04 [339]

Answer:

1.2x10⁻⁵M = Concentration of the product released

Explanation:

Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:

A = E*b*C

<em>Where A is the absotbance of the solution: 0.216</em>

<em>E is the extinction coefficient = 18000M⁻¹cm⁻¹</em>

<em>b is patelength = 1cm</em>

<em>C is concentration of the solution</em>

<em />

Replacing:

0.216 = 18000M⁻¹cm⁻¹*1cm*C

<h3>1.2x10⁻⁵M = Concentration of the product released</h3>

4 0

2 years ago