Answer:
800.0 mL.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>
M before dilution = 12.0 M, V before dilution = 100.0 mL.
M after dilution = 1.5 M, V after dilution = ??? mL.
∵ (MV)before dilution of HCl = (MV)after dilution of HCl
∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)
<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>
Answer:
The acid must be a concentrated acid
Explanation:
Ethene is prepared in the laboratory by heating ethanol with excess concentrated tetraoxosulphate VI acid at 170°C . The reactionoccursc in two stages;
1) when the ethanol and sulphuric acid are mixed in a ratio of 1:2, ethyl hydrogentetraoxosulpate VI is formed
2) The compound formed in the first step is heated in the presence of excess concentrated sulphuric acid to give ethene and sulphuric acid.
The overall reaction can be perceived as the dehydration of ethanol. The gas produced (ethene) is usually passed through sodium hydroxide solution to remove any gaseous impurities present.
concentrated sulphuric acid is used in this process since it is a good dehydrating agent.
Answer: 670K
Explanation:
Given that,
Original volume of gas V1 = 1.22 L
Original temperature T1 = 286 K
New volume V2 = 2.86 L
New temperature T2 = ?
Since volume and temperature are involved while pressure is constant, apply the formula for Charles law
V1/T1 = V2/T2
1.22 L/286 K = 2.86 L/ T2
Cross multiply
1.22 L x T2 = 286 K x 2.86 L
1.22T2 = 817.96
Divide both sides by 1.22
1.22T2/1.22 = 817.96/1.22
T2 = 670.459 K (Round to the nearest whole number as 670 K)
Thus, the temperature of the gas is 670 Kelvin
