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2 years ago

5

Number 2 & 3 please !!<3

1 answer:

Goshia [24]2 years ago

8 0

Answer:just for the ponits

Explanation:ima be a jenna

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A business owned and run by one person is a

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A business that is owned and run by one person is a sole proprietorship.

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2 years ago

What kind of chemical bond holds each oxygen atom to the phosphorus atom

MAXImum [283]

This I believe would be double covalent bonds

8 0

2 years ago

If salt is dissolved in water, water serves as the:

NemiM [27]

B. solvent

solute is the substance being dissolved

5 0

3 years ago

In which of the following cases does the reaction go to farthest to completion ?

Mashcka [7]

Answer:

$k=10^{3}$

Explanation:

k stand for equilibrium constants in terms of reaction

The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.

Consider the example below:

$R_{1}\ + R_{2}$ ⇄ $P_{1}\ + P_{2}$

where

$The\ equilibrium\ constant\ K = \frac{P_{1}P_{2}}{R_{1}R_{2}}$

For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.

5 0

2 years ago

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

3 years ago