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2 years ago

13

Help please I’ll give you brainless

1 answer:

Helen [10]2 years ago

5 0

• Liquid at room temperature
•Gas at room temperature
• Argon has approximately the same solubility in water as oxygen and is 2.5 times more soluble in water than nitrogen. Argon is colorless, odorless, nonflammable and nontoxic as a solid, liquid or gas. Argon is chemically inert under most conditions and forms no confirmed stable compounds at room temperature.

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0.467 mol NaCl are needed to

Alisiya [41]

Answer:

27.4 gram is the solution it's simple dude...

Explanation:

don't be afraid of huge question they confuse you you need not to be confused

now see simple solution

molality is denoted by m

so

m= moles of solute / mass of solvent in kg.

i hope your know the meaning of solute and solvent....

so moles are given 0.467

and molar mass is given 58.44

so just take out the gram means

by applying formula

58.44×0.467

it will give 27.4 grams simple.....

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1 year 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

How many orbitals are available in a “d” orbital configuration?

ratelena [41]

Answer:

e.)5

Explanation:

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2 years ago

Which sample of water contains particles having the highest kinetic energy

Ilya [14]

Answer:It got to be C

Explanation: 75ml of water is

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2 years ago

Read 2 more answers

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

2 years ago