Answer:
- <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>
<em />
- <em>The absolute uncertainty is: </em><u>0.008 g</u>
Explanation:
The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.
When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.
The relative uncertainty is the absolute uncertainty divided by the quantity:
- Relative uncertainty = 0.1g / 12.2 g = 0.008
The average mass of calcium is calculated using proportions, along with the molar masses:
- Molar mass of calcium: 40.078 g/ mol (from a periodic table)
- Molar mass of calcite: 100.085 g/mol (given)
Proportion:
- 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite
- x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium
So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:
- Average mass = total mass of five samples / number of samples
- Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>
So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).
The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:
- Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g
The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.
Answer:
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
Explanation:
Step 1 : Write formule of specific heat
Q=mcΔT
with Q = heat transfer (J)
with m = mass of the substance
with c = specific heat ⇒ depends on material and phase ( J/g °C)
with ΔT = Change in temperature
For this case :
Q = 1680 Calories = 7033.824 J ( 1 calorie = 4.1868 J)
m = 100.0g
c= has to be determined
ΔT = 100 - 20 = 80°C
<u>Step 2: Calculating specific heat</u>
⇒ via the formule Q=mcΔT
7033.824 J = 100g * c * 80
7033.824 = 8000 *c
c = 7033.824 /8000
c = 0,879228 J/g °C
or 0.21 Cal / g°C
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
The product is 63.8564
The correct number of S.F is 64.0000
Answer:
Cu + Cl2 → CuCl
To happen the reaction, copper metal is heated.
