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3 years ago

What do you mean by atomic number?

Chemistry

1 answer:

Allushta [10]3 years ago

4 0

Answer:

Atomic number is the number of protons in an isotope or element.

Explanation: The atomic number is the same number of protons in an element or isotope and if said isotope is neutral than it is the same number of electrons.

Example: Atomic number = 19 Protons = 19 if neutral Electrons = 19

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pogonyaev

I’ll help you 8n a bit

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2 years ago

How many H atoms in 12H2O?

Ilya [14]

Answer:

24 atoms of H or 1.4 x 10²⁵ hydrogen atoms

Explanation:

simple method

1 H₂O has atoms of hydrogen and 1 atom of oxygen

while 12H₂O will have 24 atoms of hydrogen and 12 atom of oxygen

by Avagadros number

molar mass of water H₂0=18.01528 g/mol

1 mole of H₂0 have 2 moles of Hydrogen

one mole of water= 6.02⋅10²³water molecules =1.2 x 10²⁴hydrogen atoms

12 mole of H₂O = 1.2 x 10²⁴ x 12= 1.4 x 10²⁵ hydrogen atoms

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1.4 x 10²⁵ hydrogen atoms in 12 moles of H₂O

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3 years ago

A real gas behaves more like an ideal gas when the gas molecules are

Firlakuza [10]

The answer is D, far apart and have weak attractive forces between them. The ideal gas means that the volume of molecule and the forces between them can be ignored.

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3 years ago

NaBr + CaF2 → NaF + CaBr2 What coefficients are needed to balance the chemical equation? A) 1,1,1,1 B) 1,2,1,2 C) 1,2,2,1 D) 2,1

elena-s [515]

D.
2NaBr + CaF2 --> 2NaF + CaBr2 gives you:

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This is balanced.

7 0

3 years ago

Read 2 more answers

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

3 years ago

What do you mean by atomic number? ​

What do you mean by atomic number?