All categories

2 years ago

True or false: atoms are electrically neutral; that is they do not have a charge.

1 answer:

3 0

Its true, By definition, an atom is electrically neutral<span> it has the same number of protons as it does electrons, plus some number of neutrons depending on the isotope</span>

You might be interested in

Are the particles moving faster or slower as time goes on? (Diagram F)

klasskru [66]

I YHINK ITTTSSS SHMMM BBBBB

4 0

2 years ago

Thermochemistry

max2010maxim [7]

THe Answer is 105 J

hope this helps and please brainliest, I was first :)

4 0

2 years ago

What are chemical changes of sodium hydroxide

pochemuha

Answer: Sodium + water → hydrogen gas + sodium hydroxide (aq) + heat [(aq) means “dissolved in water”. It stands for “aqueous”.] If we boil off the water, we will be able to see the sodium hydroxide crystals. The sodium hydroxide is one of the two new substances produced by this chemical change.

8 0

2 years ago

A cast fossil is formed when

Liono4ka [1.6K]

The answer is D)
<span>when the remains of a living thing are preserved and turned into rock </span>

8 0

2 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

2 years ago