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Lady_Fox [76]
3 years ago
8

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.4×1015hz?

Chemistry
1 answer:
Anika [276]3 years ago
7 0
<span>The equation you used is KE=hv-hv0, where h=6.63*10^-34 (constant). You multiply h by 1.5*10^15. Multiply h by the threshold freq of cesium (from part A). Subtract the second answer from the first answer, and you get the kinetic energy. Hope this helps.</span>
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Answer:

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Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

<em>When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ </em><em>(1)</em>

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<em>ksp = [borate²⁻] [Na⁺]²</em>

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6,00x10⁻³ mol of HCl×\frac{1molB_{4}O_{7}^{2-}}{2molHCl} = <em>3,00x10⁻³ mol of B₄O₇²⁻</em>

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Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

<em>ksp = 0,176</em>

<em></em>

I hope it helps!

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