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3 years ago

What is the oxidation number b in h3bo3?

Chemistry

1 answer:

Goryan [66]3 years ago

5 0

H is positive 1 and o is negative 2 so 3+b-6=0 and solve for b to get b=3

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Helen recorded the following data about the half-life of a radioisotope. Radioactive Decay of Radioisotope A Grams of Radioactiv

MissTica

Answer:

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Explanation:

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3 years ago

Match each item with the correct statement below.

V125BC [204]

1. LDFs

2. Intermolecular Forces

3. Intramolecular Forces

4. Linear

5. Tetrahedral

Edit: I'm new to this site and idk how to use it properly. I'm not sure about 2 and 3 currently because these forces are between molecules as well so INTERmolecular would be used twice (?)

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3 years ago

Nagine you are repeatedly lifting a weight into the air up to a height of 2 m. Assuming in each litt

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3 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

A runner competed in a 5-mile run. How many yards did she run?

vfiekz [6]

5 miles are equivalent to 8,800 yards

so b. is the correct answer. plz like and hope it helped

3 0

3 years ago