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Nimfa-mama [501]
3 years ago
7

A large cyclotron directs a beam of He++ nuclei onto a target with a beam current of 0.250 mA. (a) How many He++ nuclei per seco

nd is this? (b) How long does it take for 1.00 C to strike the target? (c) How long before 1.00 mol of He++ nuclei strike the target?
Chemistry
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

a. 7.8*10¹⁴ He⁺⁺ nuclei/s

b. 4000s

c. 7.7*10⁸s

Explanation:

I = 0.250mA = 2.5 * 10⁻³A

Q = 1.0C

1 e- contains 1.60 * 10⁻¹⁹C

But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C

(A).

No. Of charge per second = current passing through / charge

1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C

1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei

(B).

I = Q / t

From this equation, we can determine the time it takes to transfer 1.0C

I = 1.0 / 2.5*10⁻⁴ = 4000s

(C).

Time it takes for 1 mol of He⁺⁺ to strike the target =?

Using Avogadro's ratio,

1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)

Note : ions cancel out leaving the value of the answer in mols.

1.0 mol of He = 7.7 * 10⁸s

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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.
kaheart [24]

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L =  1.600M

In this case, we must set up the ICE table as shown;

              Br2(g) + I2(g) ↔ 2IBr(g)

I          0.6            1.6           0

C      -x                -x             +2x

E    0.6 - x         1.6 - x       1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 -  0.595 = 0.005

[I2] =   1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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4 0
2 years ago
**MEDAL AND FAN** what structural units make up metallic solids?
Margarita [4]
C)

Metallic solids consist of metal ions which are surrounded by delocalized electrons.
8 0
3 years ago
What is the capillary rise of ethanol in a glass tube with a 0.1 mm radius if the surface tension of ethanol is 0.032Jm2 and the
sweet [91]

Answer: The capillary rise(h) in the glass tube is = 0.009m

Explanation:

Using the equation

h = 2Tcosθ/rpg

Given

Contact angle, θ = Zero

h = height of the glass tube=?

T = surface tension = 0.032J/m^2

r = radius of the tube = 0.1mm =0.0001m

p= density of ethanol = 0.71g/cm^3

g= 9.8m/s^2

h = (2 * 0.032 * cos 0)/( 710*9.8*0.0001)

h= 0.09m

Therefore the capillary rise in the tube is 0.09m

5 0
3 years ago
Why the copper is able to replace the silver, but the reverse reaction will not result in a replacement reaction?
PSYCHO15rus [73]

Answer:

Copper is more reactive and higher in the activity series

Explanation:

Copper is able to replace the silver in a replacement reaction because it is more reactive and higher in the activity series.

Substances or ions that are higher in the activity series are typically more reactive than those ones below them.

In like manner, they are able to replace the lowers ones in a chemical reaction due to their higher reactivity potential.

A less reactive element and a lower one in the activity series cannot displace a higher one in the series.

7 0
3 years ago
Silver sulfate (Ag2SO4) is slightly soluble in water and it partly dissolves at the equilibrium according to the following balan
natka813 [3]

Answer:

a) S = 0.0152 mol/L

b) S' = 4.734 g/L

Explanation:

  • Ag2SO4 ↔ 2Ag+  +  SO42-

           S                2S            S...............in the equilibrium

  • Ksp = 1.4 E-5 = [ Ag+ ]² * [ SO42-]

a) molar solubility:

⇒ Ksp = ( 2S) ² * S = 1.4 E-5

⇒ 4S² * S = 1.4 E-5

⇒ S = ∛ ( 1.4 E-5 / 4 )

⇒ S = 0.0152 mol/L

b) solubility ( S' ) in grams per liter:

∴ Mw Ag2SO4 = 311.799 g/mol

⇒ S' = 0.0152 mol/L * ( 311.799 g/mol )

⇒ S' = 4.734 g/L

4 0
3 years ago
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