3.2 g KClO3
Explanation:
1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)
= 0.0032 mol C12H22O11
0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)
= 0.026 mol KClO3
Therefore, the minimum amount of KClO3 needed is
0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)
= 3.2 g KClO3
Answer:
The correct answer is 281.39 grams.
Explanation:
To arrive at this answer you must first keep in mind the basic equation:
<em>Q = m*Cp* ΔT</em>
Now, in order to calculate the necessary aluminum mass that absorbs 2138 J when passing from 14.83 to 23.31 ° C you must "clear" <em>m</em> of the previous equation.
This means, leave only the mass on one side of the equation, and "pass" <em>Cp</em> and <em>ΔT</em> to the other side dividing <em>Q</em>. This would look like this:
m= Q/ (Cp*ΔT)
Then, <u>you need the value of specific heat of aluminum</u> in the correct units, that is J / g ° C, the approximate value is 0.896.
ΔT is calculated by doing the mathematical operation:
23.31 °C - 14.83 °C = 8.48 °C
<em>
Finally, the values of: Q (data provided in joules), Cp (J / g ° C) and ΔT (calculated in ° C) are replaced in the last equation and the mass (in grams) is calculated resulting in 281.39 grams.</em>
there are three sientific figures
exponents decimal multiplication
Elements that are usually shiny and good conductors of heat and electricity are metals.
<h3>What are conductors?</h3>
The conductors are the metal or non metal elements which allow heat and electricity to easily pass through them.
Metals are elements with good conductivity of electric current and heat. Metals have lustre characteristics. They are shiny and bendable just like copper wire, which is a metal.
Hence, elements that are usually shiny and good conductors of heat and electricity are metals.
Learn more about conductors.
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