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1 year ago

Identify the names, charges, and location of three types of subatomic particles that make up an atom.

2 answers:

3 0

Answer:Proton (charge of +e, in the nucleus), Neutron (0 charge, in the nucleus), and Electron (charge of –e, outside the nucleus).Nov 13, 2015

Explanation: YA

vredina [299]1 year ago

3 0

Protons are positive and in the center ( nucleus)

Neutrons have no charge and are also in the center (nucleus)

Electrons are negative and orbit around the nucleus

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Assume that a cloud consists of tiny water droplets suspended (uniformly distributed,

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9.8 ms^-2 is acceleration

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3 years ago

22) The sensory cells associated with the deep layers of the epidermis are

finlep [7]

Answer:B. Merkel Cells

Explanation:

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2 years ago

Could I please get some help on this question I don’t understand .

Oksana_A [137]

Answer:

12.5 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 8 m

Final velocity (v) at 8 m above the lowest point =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 8)

v² = 0 + 156.8

v² = 156.8

Take the square root of both side

v = √156.8

v = 12.5 m/s

Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.

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2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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3 years ago

Cellular telephone use

m_a_m_a [10]

A- Radio waves, is the answer

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3 years ago

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