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3 years ago

How many atoms in:

Chemistry

1 answer:

Olin [163]3 years ago

4 0

Answer:

I think it might be:

NiO=2

CuO=4

BaO=2

Explanation:

Hope this helps.

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How do we use properties of matter in our every day lives

bearhunter [10]

For example, water. In its liquid state we drink it , in its solid state (ice ) we use it to make our drinks colder or t numbs pain when wrapped in cloth and it’s gas state ( water vapour) we steam vegetables. In conclusion many different properties of matter are used for various purposes

6 0

3 years ago

Gas is heated from 480. K to 750. K and the pressure is kept constant, what final volume would result if the original volume was

attashe74 [19]

Charles law gives the relationship between volume and temperature of gas.
It states that at constant pressure volume is directly proportional to temperature
Therefore
V/ T = k
Where V - volume T - temperature in kelvin and k - constant
V1/T1 = V2/T2
Parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
Substituting the values in the equation
267 L/ 480 K = V / 750 K
V = 417 L
Final volume is 417 L

8 0

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

Positron emission of technetium-95

NNADVOKAT [17]

Idk it’s making me answer a question in order to get help on my own

5 0

3 years ago

If the half-life of a radioactive element is 4 days, how long will it take for three- fourths of a sample of the element to deca

Jlenok [28]

Answer:

$\boxed{\text{8 da}}$

Explanation:

The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

The half-life of the element is the time it takes for half of it to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

$\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}$

$\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}$

3 0

3 years ago