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3 years ago

Please help with correct answer for the one marked :)

Chemistry

1 answer:

ehidna [41]3 years ago

5 0

Answer:

Question 4 is not clear sorry you may send another picture please

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If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele

atroni [7]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

Energy difference btween the two electronic states can be expressed as :

${ \qquad \sf \dashrightarrow \: \Delta E = h\nu}$

[ h = planks constant, ${\: \nu }$ = frequency ]

$\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}$

$\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz$

5 0

1 year 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

A certain first order reaction has a half-life of 54. 3 s. How long will it take (in s) for the reactant concentration to decrea

Maksim231197 [3]

Answer:

82.4 s

Explanation:

Find the NUMBEr of half lives...then multiply by 54.3

2.27 = 6.5 (1/2)^n

log (2.27/6.5) / log (1/2) = n = 1.52 half lives

1.52 * 54.3 = 82.4 s

8 0