Answer:
The time taken for the cross mark to disappear decreases steadily down the column.
Explanation:
Now if we look at the data provided, we will discover that the volume of the HCl was held constant while the volume of the thiosulphate was increased steadily and the volume of water decreased steadily.
Recall that a system is more concentrated when it contains less volume of water and more volume of reactants. Hence as the volume of water in the system is being reduced, the concentration of reactants is increased.
It has been established that an increase in the concentration of reactants lead to an increase in the rate of reaction. The disappearance of the cross shows the completion of the reaction between HCl and thiosulphate. The faster or slower the cross disappears, the faster or slower the rate of reaction.
Since increase in concentration of reactants increases the rate of reaction, it is observed that as the volume of the thiosulphate increases (reactant concentration increases) the cross disappears faster (rate of reactant increases). Hence as the volume of thiosulphate increases, it takes a shorter time for the cross to disappear. This implies that the time column in the table (refer to the question) will decrease steadily as the volume of thiosulphate increases.
Answer:
See explanation
Explanation:
Calcium is divalent. This means that it donates two electrons during ionic bond formation. Since chlorine atom can only accept one electron during ionic bond formation, two chlorine atoms must accept the two electrons donated by calcium.
For this purpose, each time CaCl2 is formed, there must be two chlorine atoms for each calcium atom.
<span>B. ability of a substance to transfer energy </span>
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<em>Balance</em> <em>Chemical equation:</em>
- C4H6O6 + 2NaOH ⇒ Na2C4H4O6 + 2H2O
<em>Given Data:</em>
Molarity of NaOH =0.3000
Vol. of NaOH = 21.65
<em>Solution:</em>
moles of NaOH = (0.3000 × 21.65)÷1000 = 0.0064 mol
According to balance chemical equation
NaOH : C4H6O6
2 : 1
0.0064 = (1/2) ×0.0064 = 0.0032 mol
So these 0.0032 mol of tartaric acid are prsent in 50.0 ml of solution.
So in I littre of solutions, its concentration will be as follow
M = 0.0032 × 1000/50 = 0.064 M
<em>Result: </em>
- The concentration of acid will be 0.064 M.
