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2 years ago

Explain how many protons, neutrons, and electrons can be found in an atom of carbon-12.

1 answer:

7 0

A typical atom has six protons and six neurons, and it has an atomic weight of 12 Amu. A carbon also has six electrons, but they are so small that they don’t contribute to carbons weight. An element is defined by the number of protons in its nucleus

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aniked [119]

Answer:

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2 years ago

An outdoor grill has a propane (C3H8) tank that holds 4.8 L of gas at a pressure of 5.5 atm. If all of the gas is transferred in

shtirl [24]

Answer: 0.176 atm

Explanation: Solution attached:

Use Boyle's Law to find the new pressure of the gas.

P1V1 = P2V2

Derive for P2

P2 = P1V1 / V2

= 5.5 atm ( 4.8 L ) / 150 L

= 0.176 atm

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3 years ago

The reaction of a quantity of sulfur trioxide with water gives an acid with a volume of 2 l, containing 1.8 mol of the dissolved

777dan777 [17]

Answer :]

A.)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.

<em>Correct me if i'm wrong :]</em>

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2 years ago

A covalent bond formed when 2 atoms share 6 valence electrons between them is a ______ bond

sleet_krkn [62]

COVALENT BOND IS THE BOND EXISTING BETWEEN 2 ATOMS THAT SHARE 6 ELECTRONS

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2 years ago

If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te

attashe74 [19]

Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
$pV=nRT$
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

If the supported load is W kg, then
$p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa$
Similarly,
$p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa$

$\frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}$

Because p₁/p₂ = T₁/T₂, therefore
$16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m$

The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

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3 years ago