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strojnjashka [21]
2 years ago
6

Alpha particles beamed at thin metal foil may a. pass directly through without changing direction b. be slightly diverted by att

raction to electrons c. be reflected by direct contact with nuclei d. A and C e. A, B, and C
Chemistry
1 answer:
kakasveta [241]2 years ago
6 0

Answer:

The correct option is E.

Explanation:

When alpha particles beamed at thin metal foil that time most of the particles of alpha pass through the foil with no deflection. This observation defined about two statements:

1) The diameter of the atom compared with the diameter of the nucleus.

2) The mass of an atom compared with the mass of the nucleus.

So, the correct options are:

A: Pass directly through without changing direction.

B: Be slightly diverted by attraction to electrons.

C: Be reflected by direct contact with nuclei.

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20 point, pls help. A 5.50 mole sample of a gas has a volume of 2.50 L. What would the volume be if the amount increased to 11.0
alexandr1967 [171]

Answer:

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5 0
1 year ago
Convert 5.0x10^24 molecules to liters.
makvit [3.9K]
Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!
5 0
2 years ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
An irregularly shaped solid which has a mass of 10.283g was placed in a graduated cylinder containing an inert liquid. The initi
Gwar [14]
Liquid  + Solid = 8.89 mL
V ( Solid ) = 8.89 mL - 6.26 mL = 2.63 mL
The density of the solid = m / V = 10.283 g / 2.63 mL =
= 3.9 g/mL = 3.9 g / cm³
3 0
2 years ago
The recommended daily intake of potassium ( K ) is 4.725 g . The average raisin contains 3.513 mg K . Fill in the denominators o
kondor19780726 [428]

Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

4 0
2 years ago
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