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Answer:
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Explanation:
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Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
Answer:
Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)
Explanation:
Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.
In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.
Answer:
23.8g of sodium phosphate are formed
Explanation:
Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:
3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂
<em>3 moles of sodium produce 1 mole of sodium phosphate</em>
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To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:
<em>Moles Na -Molar mass: 22.99g/mol-</em>
10g * (1mol / 22.99g) = 0.435 moles Na
<em>Moles Na₃PO₄:</em>
0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄
<em>Mass Na₃PO₄ -Molar mass: 163.94g/mol-</em>
0.145 moles Na₃PO₄ * (163.94g/mol) =
<h3>23.8g of sodium phosphate are formed</h3>