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2 years ago

Which of the following would increase the surface area of a solid so that it dissolves faster?

1 answer:

4 0

Increasing the surface area increases the rate of solubility of a solid because a larger number of molecules of the greater surface area have contact with the solvent

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3k+aici3 3kci+ai is what type of reaction

exis [7]

3K + AlCl3 --------> 3 KCl + Al
element compound compound element

This is single replacement (displacement) reaction.

3 0

2 years ago

What is the molar mass of <br> C4H9, 114.17 g/mol

mihalych1998 [28]

Answer:

4 x 12 +9 x 1= 56

Explanation:

I do not know what the 114.17 g/mol comes from

6 0

3 years ago

Read 2 more answers

The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 9.00 g

timofeeve [1]

Answer:

20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

Explanation:

Heat is being consumed during vaporization and heat is being released during condensation.

To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.

Molar mass of water = 18.02 g/mol

Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.

So, to vaporize 9.00 g of water, $(\frac{40.66}{18.02}\times 9.00)kJ$ of heat or 20.3 kJ of heat is being consumed

As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

5 0

3 years ago

What is the variable that is changed by

denpristay [2]

Answer:

b. independent/manipulated variable

Explanation:

Independent/manipulated variable - refers to the variable that is changed by the scientist or an experimenter. Only one variable that is independent is required to ensure a fair test in an excellent experiment. As the independent variable is being changed by an experimenter or scientist, data is being recorded simultaneously as they are collected.

3 0

1 year ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago