Answer:
option C is correct = 1.14 × 10²² molecules of CO₂
Explanation:
Given data:
Number of moles of CO₂ = 0.0189 mol
Number of molecules = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
For given question:
1 mole of CO₂ = 6.022 × 10²³ molecules of CO₂
0.0189 mol of CO₂ × 6.022 × 10²³ molecules of CO₂ / 1mol
1.14 × 10²² molecules of CO₂
Thus, option C is correct.
Answer:
Compound B is ionic. The electronegativity difference is 2.2, which can be determined by subtracting the electronegativity of Element Y from that of Element Z. Electronegativity differences greater than 1.7 indicate ionic bonds.
Hope that helps.
Answer:

<em><u>Step by step explanation</u></em><em><u>:</u></em>
<em>Given:</em>
Mass of given sample (m) = 2.50 g
Initial temperature (T1) = 25°C
Final temperature (T2) = 20°C
Heat Energy Q = 12 cal
<em>T</em><em>o </em><em>find:</em>
<em>
</em>
<em>Solution</em><em>:</em>
We know that,
<em>Specific</em><em> </em><em>heat</em> <em>of </em><em>any </em><em>substance </em><em>is </em><em>directly</em><em> </em><em>proportional</em><em> </em><em>to </em><em>the </em><em>mass </em><em>and </em><em>change </em><em>in </em><em>temperature.</em>
Represented by equation,

Where,
<em>Q = Heat Energy</em>
<em>m = mass of given sampl</em><em>e</em>
<em>c = specific heat</em>
<em>∆T = change in </em><em>temperature</em>
Substituting corresponding values,
<em>
</em>
We also know that,

multiplying above answer by 4.184,

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4.21 x 10^-4
This should be it