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sergij07 [2.7K]
4 years ago
13

Bob is investigating whether the density of water changes with temperature. Which of the following variables should he change in

his experiment?
temperature of the water

type of water used (bottled water or tap water)

volume of the water used

mass of the water used
Chemistry
1 answer:
LekaFEV [45]4 years ago
7 0

Answer:

temperature of the water

Explanation:

Density is defied as mass divided by the volume. To investigate how the density of water change with temperature, Bob have the change the temperature and read the volume and mass of the investigated sample.

Even if there is a different salt content between booted water or tap water and have an influence on the density (compared with pure water), this difference is not so big so the change in density with temperature can be determined. Considering that the experiment do not require extreme accuracy,  any type of water may be used (bottled water of tap water).

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olga2289 [7]

Answer:

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Explanation:

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5 0
4 years ago
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change
MrRa [10]

Answer:

-54 kJ/mol

Explanation:

Given that:

A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol

i.e

50 ml of 1.0 M HCl +  50 ml of 1.0 M NaOH -----> -54 kJ/mol

If he repeat the same experiment with :

100 ml of 1.0 M HCl + 100 ml of 1.0 M NaOH. ------> ????

From The experiment; the molar enthalpy of change of the acid-base neutralization reaction will be -54 kJ/mol

This is because : The second reaction requires 50 ml in order to neutralize the reaction, then the remaining 50 ml will be excess, Hence, there is no change in the enthalpy of the reaction.

Similarly; we can assume that :

In the first reaction;  P moles of  is used to liberate Q kJ heat ; then  the change in molar enthalpy will be Q/P (kJ/mol).

SO; when he used 100 ml ;

then the amount of moles used is double, likewise the heat liberated will be doubled ;

So;

2P moles is used to liberate 2Q kJ heat ;

2P/2Q = Q/P ( kJ/mol) = -54 kJ/mol

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