2H2 + O2 ---> 2H2O

1 answer:

7 0

2 H2 + O2 ----------- 2 H2O

2 mol H2 ----------- 1 mol O2
27.4 mol H2- ------ x mol O2

2x = 27.4

x = 27.4 / 2

x = 13.7 moles O2

answer b

hope this helps!.

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Does the reaction of a main-group metal oxide in water produce an acidic solution or a basic solution? Write a balanced equation

Fynjy0 [20]

<u>Answer:</u> The main group metal produce a basic solution in water and the reaction is $MO+H_2O\rightarrow M(OH)_2$

<u>Explanation:</u>

Main group elements are the elements that are present in s-block and p-block.

The metals that are the main group elements are located in Group IA, Group II A and Group III A.

Oxides are formed when a metal or a non-metal reacts with oxygen molecule. There are two types of oxides which are formed: Acidic oxides and basic oxides.

Acidic oxides are formed by the non-metals.
Basic oxides are formed by the metals.

When a metal oxide is reacted with water, it leads to the formation of a base.

The general formula of the oxide formed by Group II-A metals is 'MO'

The chemical equation for the reaction of metal oxide of Group II-A and water follows:

$MO+H_2O\rightarrow M(OH)_2$

Hence, the main group metal produce a basic solution in water and the reaction is $MO+H_2O\rightarrow M(OH)_2$

4 0

3 years ago

Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3, what is the sample’s volume?

IRINA_888 [86]

Density (p) is defined as the mass (m) per unit volume (v) or:

p = m/v

Using this relationship, the volume is:

v = m/p

Using the given values of mass of 80 grams and density of 8 g/cm3, the sample volume is:

v = 80 grams/8 grams/cm3
v = 10 cm3

The final answer is 10 cm3.

7 0

4 years ago

A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temp

Novosadov [1.4K]

Answer:

1) Exothermic.

2) $Q_{rxn}=-8580J$

3) $\Delta _rH=-121.0kJ/mol$

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

$Q_{rxn}=-m_{Total}C(T_2-T_1)\\\\Q_{rxn}=-(300g+10.7g)*4.184 \frac{J}{g\°C} (28.6\°C-22.0\°C)\\\\Q_{rxn}=-8580J$

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case LiCl, we proceed as follows:

$\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J} \\\\\Delta _rH=-121.0kJ/mol$