In a typical double displacement reaction, you would have a total of two products (AB + CD —> AD + BC).
Remember that density refers to the "mass per unit volume" of an object.
So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.
In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is
As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:
density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml
The effect on density would be that it would erroneously appear to be greater
Hope this helps!
Good luck
Chemical reactions are basically divided into two major classes depending on whether the reaction lose energy or gain energy from the environment during the course of the reaction. The two classes of reaction are exothermic and endothermic reaction.
An exothermic reaction is a type of reaction in which the reaction system lose energy to the environment and thus, the energy content of the reactants is more than that of the product formed. Because of this, the enthapyl change of an exothermic reaction is always negative.
An endothermic reaction is a type of reaction in which the reaction system absorb energy from the environment. Thus, the energy contents of the products is always higher than that of the reactants and the enthapyl change of the reaction is always positive. During the course of the reaction, the reaction container is usually cold to the touch because energy is been absorbed from the environment.
i believe it would be B '' tetrahedral compound ''
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
