Reaction of sodium with water
Sodium metal reacts rapidly with water to form a colourless solution of sodium hydroxide (NaOH) and hydrogen gas (H2). The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic. During the reaction, the sodium metal may well become so hot that it catches fire and burns with a characteristic orange colour. The reaction is slower than that of potassium (immediately below sodium in the periodic table), but faster than that of lithium (immediately above sodium in the periodic table).
2Na(s) + 2H2O → 2NaOH(aq) + H2(g)
Answer:
D. To ensure the cooling process is not affected by surrounding temperature
Explanation:
The conical flask acts as a <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>j</u><u>a</u><u>c</u><u>k</u><u>e</u><u>t</u><u>.</u>
<u>Answer:</u> The red litmus paper turns blue on dipping in NaOH solution.
<u>Explanation:</u>
Litmus paper is the indicator that detects the nature of the solution, whether it is acidic or basic.
There are 2 types of litmus paper:
- <u>Red litmus paper:</u> This paper will turn blue if it is dipped in basic solution and will remain as such if it is dipped in acidic solution.
- <u>Blue litmus paper:</u> This paper will turn red if it is dipped in acidic solution and will remain as such if it is dipped in basic solution.
NaOH is a strong base, so when a red litmus paper is dipped in the beaker having necessary amount of NaOH, the red litmus paper turns into blue.
Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,

where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:


Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration
The answer would actually be false. I just took the test.