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murzikaleks [220]
3 years ago
8

how does the structure of atoms (i.e. protons, neutron, and electrons) affect the electric charge of an object or substance.

Chemistry
1 answer:
pickupchik [31]3 years ago
8 0

Electrons contribute greatly to the atom's charge, as each electron has a negative charge equal to the positive charge of a proton. Scientists define these charges as “+1” and “-1. ” In an uncharged, neutral atom, the number of electrons orbiting the nucleus is equal to the number of protons inside the nucleus.:

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During a chemical change items are lost or gained to make new substance true or false​
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Answer:

If you meant Atoms then the answer is false

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Oxygen and hydrogen are both elements that are found as gases at room temperature. When oxygen combines with hydrogen, they prod
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A

Explanation:

has properties that are different from the original substances.

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what is the predicted order of fist ionization energies from highest to lowest for beryllium, calcium, magnesium, and stronium?
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Read 2 more answers
(01.04 MC)
iogann1982 [59]
I turned in this exact assignment today haha

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6 0
3 years ago
A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern
Arada [10]

<u>Answer:</u> The volume of the vessel is 0.1542m^3 and total internal energy is 162.0 kJ.

<u>Explanation:</u>

  • To calculate the volume of water, we use the equation given by ideal gas, which is:

PV=nRT

or,

PV=\frac{m}{M}RT

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g    (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of container = 60^oC=[60+273]K=333K

Putting values in above equation, we get:

700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L

Converting this value into m^3, we use the conversion factor:

1m^3=1000L

So, \Rightarrow (\frac{1m^3}{1000L})\times 154.21L

\Rightarrow 0.1542m^3

  • To calculate the internal energy, we use the equation:

U=\frac{3}{2}nRT

or,

U=\frac{3}{2}\frac{m}{M}RT

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g  (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = 8.314J/K.mol

T = temperature = 60^oC=[60+273]K=333K

Putting values in above equation, we get:

U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is 0.1542m^3 and total internal energy is 162.0 kJ.

6 0
3 years ago
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