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3 years ago

An atom's nucleus contains three protons and three neutrons. What is the overall charge of the nucleus?

2 answers:

8 0

If the number of electrons is not given, your answer would be +3.

Explanation: Protons have positive charge, electrons have negative charge, and neutrons have no charge. Therefore, having three protons and 3 neutrons would give you a net +3 charge.

Hope this helps!

lana66690 [7]3 years ago

3 0

Answer:

Explanation:

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Converts 800kg into ng

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The elements Lithium (Li), Boron (B), and Oxygen (O) are all in the same _________ on the periodic table.

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Answer:

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3 years ago

A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is

makvit [3.9K]

Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

$\frac{V}{T} =k$

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

$\frac{V1}{T1} =\frac{V2}{T2}$

<h3>Temperature of the gas in this case</h3>

In this case, you know:

P1= 1800 psi
V1= 10 L
T1= 20 C= 293 K (being 0 C= 273 K)
P2= 1800 psi
V2= 6 L
T2= ?

You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

$\frac{10 L}{293 K} =\frac{6 L}{T2}$

Solving:

$\frac{10 L}{293 K} T2=6 L$

$T2=\frac{6 L}{\frac{10 L}{293 K} }$

The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

brainly.com/question/4147359?referrer=searchResults

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3 years ago

A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what

KIM [24]

Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

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3 years ago

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