Answer:
True
Explanation:
The length of covalent bond depends upon the size of atoms and the bond order.
a balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.
Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
Answer: A
Explanation:
Because I just answered it and it’s right. ( Forces between electron pairs push the atoms apart.
To do this problem, we must first look at the balanced chemical equation for the decomposition of potassium chlorate:
<span>2KClO3 --> 2KCl + 3O2 </span>
<span>We can take the given amount of grams, and use the molar mass of KClO3 to convert to moles. Then, we can use the stoichiometric ratios to relate moles of KClO3 to moles of O2. </span>
<span>(39.09)+(35.45)+(3*15.99)= 122.51 g/ mol = molar mass of KClO3 </span>
<span>45.8 g KClO3/ 122.51 g/ mol KClO3 = .374 moles KClO3 </span>
<span>.374 mol KClO3 *(3 moles O2/2 mol KClO3)= .560 moles O2 </span>
<span>Once we have moles of O2, we can convert to grams of O2. </span>
<span>(2*15.99)= 31.98 g/mol = molar mass of O2 </span>
<span>(.560 moles O2) (31.98 g/mol)= 17.91 g O2 </span>
<span>Hope this helps :)</span>
