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3 years ago

Which electromagnetic waves have the hugest energy?

Chemistry

1 answer:

Blababa [14]3 years ago

5 0

Answer:

X-Rays

Explanation:

X-rays have shorter wavelengths (higher energy ) than UV waves and, generally, longer wavelengths (lower energy) than gamma rays.

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NEED HELP ASAP!!!!!!!!!!!!!!!!!

Leya [2.2K]

Answer:

Decomposition.

Explanation:

It is decomposition, result is 2H2+O2, hope it is correct.

5 0

3 years ago

Is anyone good with science?

valkas [14]

Answer:

Yes

Explanation:

4 0

3 years ago

When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit

pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

$\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)$ .

How many moles of this process?

Relative atomic mass from a modern periodic table:

K: 39.098;
N: 14.007;
O: 15.999.

Molar mass of $\text{KNO}_3$ :

$M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of the process = Number of moles of $\text{KNO}_3$ dissolved:

$\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}$ .

What's the enthalpy change of this process?

$Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}$ for $0.212162\;\text{mol}$ . By convention, the enthalpy change $\Delta H$ measures the energy change for each mole of a process.

$\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}$ .

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0

3 years ago

How many moles of NO3 are there in 6.2g?

Nataliya [291]

6.2 divided by the mass number of NO3 is 0.1 moles. Anytime your trying to find the moles of something you divide it by the mass number of your combined elements.

3 0

3 years ago

Read 2 more answers

The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c

Ivahew [28]

Explanation:

(a) The given data is as follows.

B = $2.180 \times 10^{-18} J$

Z = 4 for Be

Now, for the first excited state $n_{f}$ = 2; and $n_{i} = \infinity$ if it is ionized.

Therefore, ionization energy will be calculated as follows.

I.E = $\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})$

= $8.72 \times 10^{-18} J$

Converting this energy into kJ/mol as follows.

$8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol$

= 5249 kJ/mol

Therefore, the ionization energy of the $Be^{3+}$ ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

$\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}$

$\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}$

$\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}$

= $303.7 \times 10^{-10} m$

or, = $303.7^{o}A$

Therefore, wavelength of light given off from the $Be^{3+}$ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels $303.7^{o}A$ .

5 0

3 years ago