V=0 m/s, u=?, a = 9.8 m/s² and s = 0.8 m
u²=2×9.8×0.8=15.68
<span>u=3.959 ≈ 3.96 m/s
</span>
Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
<span>2NaCN + (1)H2SO4 → Na2SO4 + 2HCN
</span><span>The coefficient of sulfuric acid is 1.</span>
Answer:
1
Explanation:
4 HBr + O2 → 2H 20 + 2Br 2
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