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2 years ago

Why did most of the alpha particles go straight through the gold foil in Rutherford's experiment?

Chemistry

1 answer:

andre [41]2 years ago

4 0

Answer:

The fact that most alpha particles went straight through the foil is because the atom is mostly empty space.

Those that passed straight through did so because they didn't encounter any nuclei.

Explanation:

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Akimi4 [234]

Answer:

genes (c)

Explanation:

since traits are passed down genetically, so her mother or father must have one (or even one of her grandparents)

hope this helps!:)

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2 years ago

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Convert 18g of Li into atoms.

gizmo_the_mogwai [7]

The full decimal is 2.59328...
When you round up the answer is 2.6 atoms of Li

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3 years ago

For most substances,the distance between particles is furthest when the substance is?

valentinak56 [21]

Im not positive but i think it is When it is a gas?

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3 years ago

1. Calculate the [H] in a solution that has a pH of 9.88.

Olenka [21]

The answer to the question is D.

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3 years ago

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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago