Answer: The transition elements are in the d-block, and in the d-orbital have valence electrons. They can form several states of oxidation and contain different ions. Inner transition elements are in the f-block, and in the f-orbital have valence electrons.
Explanation:
Reaction:
<span>HCl + NaOH ---> NaCl + H2O
</span><span>1 mole of HCl = 36,5 g
</span><span>1 mole of NaOH = 40g
</span><span>so, according to the reaction:
</span><span>1 mol HCl = 1 mol NaOH
</span>so, we need > 36,5 g HCl (<u>hydrochloric acid</u><span>)
</span><u>
answer: 36,5 g HCl (hydrochloric acid)
</u><span> ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
</span><span>next question.
</span><span>
1 mole of NaCl = 58,5 g
</span><span>1 mole of H2O = 18g
</span>
so, according to the reaction:
1 mole of HCl (36,5 g) <span>----------------- - 1 mole of NaCl (58,5 g)
</span><span>(the same for NaOH)
i
</span>1 mole of HCl<span> (36,5 g) ------------------ 1 mole of H2O (18 g)
</span>(the same for NaOH)
<span>so, this reaction is stechiometric
</span><u>
answer: 58,5 g NaCl i 18g H2O</u>
Answer:
F = 800 N
Explanation:
Given data:
Mass = 80 Kg
Acceleration = 10 m/s²
Force = ?
Solution:
Formula:
<em>F = m × a
</em>
F = force
m = mass
a = acceleration
Now we will put the values in formula:
<em>F = m × a
</em>
F = 80 kg <em>× </em>10 m/s²
F = 800 kg.m/s²
kg.m/s² = N
F = 800 N
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.
so this is right answer
For reasons that are unclear, no eukaryotic enzymes can break the triple bond of N2. The reduction of N2 to NH3 (nitrogen fixation) is limited to prokaryotes and is catalysed by nitrogenase. Since most of the nitrogen entering the biosphere (around 100 million metric tonnes of N2 per annum) does so through nitrogenase activity (lightning contributes about 10%), those plants that associate with nitrogen-fixing bacteria have a significant selective advantage under conditions of limiting nitrogen.