<span>B. electron = negative; proton = positive; neutron = no charge </span>
Answer:The new volume is 5mL
Explanation:
The formular for Boyles Law is; P1 V1 = P2 V2
Where P1 = 1st Pressure V1 = First Volume
P2 = 2nd Pressure V2 = Second Volume
From the question; P1 = 5atm, V1 = 10ml
P2 = 2 x P1 (2 x 5) = 10 atm V2 =?
Using the Boyles Law Formular; P1 V1 = P2 V2, we make V2 the subject of formular; P1 V1/ P2 = V2
∴ 5 x 10/ 10 = 5
∴ V2 = 5mL
Answer:
1.552 moles
Explanation:
First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.
This is illustrated below:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 = 22.1g
Mole of C8H18 =..?
Number of mole = Mass /Molar Mass
Mole of C8H18 = 22.1/144
Mole of C8H18 = 0.194 mole
From the balanced equation above,
2 moles of C8H18 produced 16 moles of CO2.
Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.
Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned
Answer:
The answer to your question is: letter E
Explanation:
Normally, the correct order of boiling points is:
Alcohols > Ketones > Ether > Alkane
Then
A. n-butane < 1-butanol < diethyl ether < 2-butanone
B. n-butane < 2-butanone < diethyl ether < 1-butanol
C. 2-butanone < n-butane < diethyl ether < 1-butanol
D. n-butane < diethyl ether < 1-butanol < 2-butanone
E. n-butane < diethyl ether < 2-butanone < 1-butanol
(- 1°C) < 34.6°C < 79.64°C < 117.7°C
Part A
75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF
This combination will form a buffer.
Explanation
Here, weak acid HF and its conjugate base F- is available in the solution
Part B
150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
This combination cannot form a buffer.
Explanation
Here, moles of HF = 0.15 x 0.1 = 0.015 moles
Moles of HCl = 0.135 x 0.175 = 0.023
Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution
Part C
165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
This combination will form a buffer.
Explanation
Moles of HF = 0.165 x 0.1 = 0.0165 moles
Moles of KOH = 0.135 x 0.05 = 0.00675 moles
Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer
Part D
125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
This combination will form a buffer
Explanation
Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer
Part E
105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
This combination will form a buffer
Explanation
Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles
Moles of HCl = 0.095 x 0.1 = 0.0095 moles
Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
