Answer:
CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)
Explanation:
We have the reactants as calcium chloride and potassium carbonate. Recall that we are expecting that the reaction will yield a precipitate. We must keep that in mind as we seek to write its balanced chemical reaction equation.
So we now have;
CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)
Recall that the rule of balancing chemical reaction equation states that the number of atoms of each element on the right side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation.
A baseline for experimental investigation is provided by an hypothesis. This is a must before conducting experiments. Also, it is the hypothesis that is being proved by doing the experiments. So, hypothesis is very important in research studies. Hope this answers the question.
Mass using grams because of the balance scale is evenly weighted not from springs and gravity like a normal scale.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
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