You are given
200 grams of H2O(s) at an initial temperature of 0°C. you are also given the
final temperature of water after heating at 65°C. You are required to get the
total amount of heat to melt the sample. The specific heat capacity, cp, of
water is 4.186 J/g-°C. Let us say that T1 = 0°C and T2 = 65°C. The equation for
heat, Q, is
Q = m(cp)(T2-T1)
Q = 200g(4.186
J/g-°C )(65°C - 0°C)
<u>Q =
54,418J</u>
The balanced chemical reaction is
<span>2al + 3cl2 = 2alcl3
To determine the maximum amount of product, we need to determine which is the limiting reactant. Then, use the initial amount of that reactant to calculate the amount of the product that would be produced. We do as follows:
7 mol Al (3 mol Cl2 / 2 mol Al) = 10.5 mol Cl2
8 mol Cl2 ( 2 mol Al / 3 mol Cl2) = 5.3 mol Al
Thus, it is Cl2 that is the limiting reactant.
8 mol Cl2 ( 2 mol AlCl3 / 3 mol Cl2) = 5.3 moles of AlCl3 is produced</span>
Answer:
look at the graph
Explanation:
We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.
Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.
So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)
Answer:
The answer to your question is: 17.26% of carbon
Explanation:
Data
CxHy = 0.2121 g
BaCO₃ = 0.6006 g
Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g
Reaction
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Process
1.- Find the amount of carbon in BaCO₃
197 g of BaCO₃ --------------- 12 g of Carbon
0.6006 g ---------------- x
x = (0.6006 x 12) / 197
x = 0.0366 g of carbon
2.- Calculate the percentage of carbon in the organic compound
0.2121 g of organic compound --------------- 100%
0.0366g -------------- x
x = (0.0366 x 100) / 0.2121
x = 17.26%
2.5LX2 M=5 moles
As Molarity X Volume= Total Number of Moles
