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3 years ago

What makes up a film badge?

Chemistry

2 answers:

Inessa [10]3 years ago

8 0

Film badges is a personal dosimeter used for monitoring cumulative radiation dose due to ionizing radiation.

A panel or layers of photographic film covered with black paper and exposed to air ,salt,and other chemicals on clear acetate

Sliva [168]3 years ago

4 0

Answer:

layers of photo graphic film covered in black paper

Explanation:

The badge consists of two parts: photographic film and a holder.[1] The film emulsion is black and white photographic film with varying grain size to affect its sensitivity to incident radiation such as gamma rays, X-rays and beta particles.

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strojnjashka [21]

Answer:

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2 years ago

Definition of periodic trends

Dmitriy789 [7]

Answer:

.Periodic trends are specific patterns in the properties of chemical elements that are revealed in the periodic table of elements. Major periodic trends include electronegativity, ionization energy, electron affinity, atomic radii, ionic radius, metallic character, and chemical reactivity.

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3 years ago

If 100.0 grams of na3n decompose to form sodium and nitrogen, how many moles of sodium are formed? write and balance equation be

Yakvenalex [24]

The balanced equation is:
$2Na_3N-\ \textgreater \ 6Na + N_2$

Then proceed with the following equations.

$100g Na_3N*(\frac{1molNa_3N}{82.98gNa_3N})*(\frac {6mol Na}{2molNa_3N})*(\frac {22.99gNa}{1molNa})=83.12gNa$

The answer is $83.12gNa$ .

7 0

3 years ago

Read 2 more answers

Consider the reaction.

Alex_Xolod [135]

Note the signs of equilibrium:-

Reaction don't procede forward or backward
Concentration of products and reactants remains same .

Concentration of A is 2M then concentration of B should be same .

So equilibrium constant K is 1

$\\ \rm\rightarrowtail K=\dfrac{[Products]^a}{[Reactants]^b}$

[B]=[A]^2
[B]=2^2
[B]=4M

8 0

2 years ago

Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the

expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

Volume of $Ca(OH)_2$ = 14.5 mL

Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

So, in 14.5mL. the mass of $Ca(OH)_2$ present will be = $\frac{0.185}{100}\times 14.5=0.0268g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = $2.50\times 10^{-3}$

Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

5 0

3 years ago