Pressure: when the external pressure is:
less than one atmosphere, the boiling point of the liquid is lower than its normal boiling point.
equal to one atmosphere, the boiling point of a liquid is called the normal boiling point.
greater than one atmosphere, the boiling point of the liquid is greater than its normal boiling point.
Answer:
250000 μL
Explanation:
If 1 L = 1000 mL
Then X L = 250 mL
X = (1 × 250) / 1000 = 0.25 L
Now we can calculate the number of microliters (μL) in 0.25 L:
if 1 μL = 10⁻⁶ L
then X μL = 0.25 L
X = (1 × 0.25) / 10⁻⁶ =250000 μL
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
Answer:
0.098 moles
Explanation:
Let y represent the number of moles present
1 mole of Ba(OH)₂ contains 2 moles of OH- ions.
Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.
To get the y moles, we then do cross multiplication
1 mole * y mole = 2 moles * 0.049 mole
y mole = 2 * 0.049 / 1
y mole = 0.098 moles of OH- ions.
1 mole of OH- can neutralize 1 mole of H+
Therefore, 0.098 moles of HNO₃ are present.
The question is incomplete, complete question is :
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?
Structure is given in an image?
Answer:
There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Explanation:
Total numbers of carbon = 10
Number of primary carbons that is carbon joined to just single carbon atom = 6
Number of secondary carbons that is carbon joined to two carbon atoms = 1
Number of tertiary carbons that is carbon joined to three carbon atoms = 2
Number of quartenary carbons that is carbon joined to four carbon atoms = 1
So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.