Answer:
The correct option is the last option
Explanation:
The chemical reaction provided in the question is a double-displacement reaction which is an exothermic reaction (which is the reason for the release of heat). An example of a double displacement reaction and exothermic reaction is the neutralization reaction illustrated below.
HCl + NaOH ⇒ NaCl + H₂O
From the law of conservation of energy, energy can neither be created nor destroyed, hence the total energy in a given system should ordinarily be the same (in the reactants and products), however <u>when energy is released in a reaction (as in the case with an exothermic reaction), it shows there are more bond energy in the reactants than in the products and it is the excess energy that is been released into the atmosphere.</u>
Answer:
Explanation:
mass of the reactant = mass of the product
15.31 + 1.50 = 16.81g
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
Evaporation is the process by which a liquid changes to a gas at temperatures lower than its boiling point
BUT
Boiling is the process by which a substance changes from a liquid to a gas at the boiling temperature of the substance.
Hope the photo helps too
1. 70 with 1 sig fig
2. 1000 with 1 sig fig
3. 4040 with 3 sig fig
4. I’m not certain if that second part is 2.6X10-12 but if so, the answer is 20 with 1 sig fig