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sergiy2304 [10]
3 years ago
5

Four properties of water are listed. It has high thermal conductivity. It reacts with metals to form metal hydroxide and hydroge

n gas. When heated to 100 °C, water evaporates. It decomposes to form hydrogen and oxygen. Which are the chemical properties of water? 1 and 2 1 and 3 2 and 3 2 and 4
Chemistry
2 answers:
Gnom [1K]3 years ago
7 0
2nd and 4th properties are chemical properties
erica [24]3 years ago
5 0

Answer:

The chemical properties are then number 2 and 4:

  • It reacts with metals to form metal hydroxide and hydrogen gas.
  • It decomposes to form hydrogen and oxygen.

Justification:


The chemical properties can only be observed when the substance undergoes a chemical change, which is that the substance is transformed into one or more different substances.  As result of a chemical change, the composition of the former substance changes, braking some chemcial bonds and formng new ones.


The number 1 states that the water reacts with metals to form metal hydroxide, meaning that the water is transformed into a new substance, the composition will not be H₂O andy more, since the atoms will be combined with atoms of different elements. Same, can be told about the number 2, which states that hydrogen and oxygen were obtained when water decomposed.


On the other hand, the thermal conductivity of water and the boiling point (the temperature at which the water evaporates) can be measured without changing the composition of the water, so those are physical properties.

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1. A sample of gas has an initial volume of 25 L and an initial pressure of 123.5 kPa. If the pressure changes to
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Answer:

\large \boxed{\text{17.mL}}

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{123.5 kPa}\qquad & V_{1} &= & \text{25 L} \\p_{2}& =& \text{179.9 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:  

\begin{array}{rcl}123.5 \times 25 & =& 179.9V_{2}\\3088 & = & 179.9V_{2}\\V_{2} & = &\dfrac{3088}{179.9}\\\\& = &\textbf{17 L}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{17 L}}

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3 years ago
Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating
kvv77 [185]

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = \frac{1}{1}\times 0.0295=0.0295moles of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g

To calculate the percentage yield of iodo-salicylamide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%

Hence, the percent yield of the reaction is 68.68%.

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