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3 years ago

Geiger counters and scintillation counters differ in Blank .

Chemistry

2 answers:

AfilCa [17]3 years ago

4 0

Answer:

Geiger counters can detect all kinds of radiation, while scintillation counters can only detect ionizin radiation

butalik [34]3 years ago

3 0

The Geiger Counter. Geiger counters are used to detect radioactive emissions, most commonly beta particles and gamma rays. The counter consists of a tube filled with an inert gas that becomes conductive of electricity when it is impacted by a high-energy particle.

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Can someone please please help me out ?

GarryVolchara [31]

Answer:

4.823 x 10^-19 J

Explanation:

Energy is calculated by E = hv where h - Planck's constant in joule.s

v - frequency.

in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s

-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec

now we can simply use Planck's constant in E=hv =

(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.

6 0

2 years ago

If the reaction is at dynamic equilibrium at 500 K, which statement applies to the given chemical system?

Alenkinab [10]

The correct option is this: THE CONCENTRATION OF THE PRODUCTS AND THE REACTANTS DO NOT CHANGE.

A reversible chemical reaction is said to be in equilibrium if the rate of forward reaction is equal to the rate of backward reaction. At this stage, the concentrations of the products and the reactants remain constant, that is, there is no net change in the concentration even though the reacting species are moving between the forward and the backward reaction.

8 0

3 years ago

Read 2 more answers

Hurry I need the answer asap

olga nikolaevna [1]

Answer:

Heat or electricity in the atmasphere

Explanation:

4 0

2 years ago

What is the molarity of a solution composed of 5.85 g of potassium iodide, KI, dissolved

Troyanec [42]

Answer:

0.282 M

General Formulas and Concepts:

Chemistry - Solutions

Reading a Periodic Table
Using Dimensional Analysis
Molarity = moles of solute / liters of solution

Explanation:

Step 1: Define

5.85 g KI

0.125 L

Step 2: Identify Conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of I - 126.90 g/mol

Molar Mass of KI - 39.10 + 126.90 = 166 g/mol

Step 3: Convert

$5.85 \ g \ KI(\frac{1 \ mol \ KI}{166 \ g \ KI} )$ = 0.035241 mol KI

Step 4: Find Molarity

M = 0.035241 mol KI / 0.125 L

M = 0.281928

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.281928 M ≈ 0.282 M

7 0

3 years ago

If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w

irinina [24]

Answer:

4.13 g

Explanation:

Data Given:

Amount of CO₂ Produced = 15 g

Amount of Oxygen = 11 g

Amount of Carbon used = ?

Solution:

Suppose Carbon dioxide (CO₂) is formed by the reaction of carbon and oxygen then the reaction will be as below

C + O₂ -------------> CO₂

1 mol 1 mol 1 mol

we come to know from the above reaction that

1 mole of carbon react with 1 mole of oxygen to produce 1 mol of carbon dioxide.

molar mass of C = 12 g/mol

molar mass of O₂ = 32 g/mol

molar mass of CO₂ = 12 + 2(16) = 44 g/mol

if we represent mole in grams then

C + O₂ -------------> CO 1 mol (12 g/mol) 1 mol (32 g/mol) 1 mol (44 g/mol)

C + O₂ -------------> CO₂

12 g 32 g 44 g

So,

we come to know that 32 g of Oxygen combine with 12 g of oxygen produce 44 g CO₂

So now how much of Carbon will be combine with 11 g of oxygen

apply unity formula

32 g of O₂ ≅ 12 g of C

11 g of O₂ ≅ g of C

by doing cross multiplication

g of C = 12 g x 11 g / 32 g

g of C = 132 g / 32 g

g of C = 4.13 g

So,

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

to check this answer

we use the above information

12 g of C ≅ 44 g of CO₂

4.13 g of C ≅ g of CO₂

by doing cross multiplication

g of CO₂ = 44 g x 4.13 g / 12 g

g of CO₂ = 15g

So it is confirmed that

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

4 0

3 years ago