The present carbon skeleton in an ether is C-C-C-O-C-C-C because ethers conatins C-O-C as functional group
Missing in your question:
Picture (1)
when its an open- tube manometer and the h = 52 cm.
when the pressure of the atmosphere is equal the pressure of the gas plus the pressure from the mercury column 52 Cm so, we can get the pressure of the gas from this formula:
P(atm) = P(gas) + height (Hg)
∴P(gas) = P(atm) - height (Hg)
= 0.975 - (520/760)
= 0.29 atm
Note: I have divided 520 mm Hg by 760 to convert it to atm
Picture (2)
The pressure of the gas is the pressure experts by the column of mercury and when we have the Height (Hg)= 67mm
So the pressure of the gas =P(atm) + Height (Hg)
= 0.975 + (67/ 760) = 1.06 atm
Picture (3)
As the tube is closed SO here the pressure of the gas is equal the height of the mercury column, and when we have the height (Hg) = 103 mm. so, we can get the P(gas) from this formula:
P(gas) = Height(Hg)
= (103/760) = 0.136 atm
Answer:
la particula que queda es h2o
Explanation:
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.
I guess it is the second one but you missed the state symbols.
