Potential differences between the ends of a circuit. Electrons flow from high potential to low potential.
Answer:
Final T = 64.0°C.
Explanation:
- The amount of heat absorbed by Na (Q) can be calculated from the relation:
<em>Q = m.c.ΔT.</em>
where, Q is the amount of heat absorbed by Na (Q = 1840 J),
m is the mass of Na (m = 68.0 g),
c is the specific heat capacity of Na (c = 1.23 J/g °C),
ΔT is the temperature difference (final T - initial T) (ΔT = final T - 42.0°C).
∵ Q = m.c.ΔT.
∴ (1840 J) = (68.0 g)(1.23 J/g °C)(final T - 42.0°C)
(final T - 42.0°C) = (1840 J)/(68.0 g)(1.23 J/g °C) = 22.0°C.
<em>∴ final T</em> = 22.0°C + 42.0°C = <em>64.0°C.</em>
Answer:
1.0M HCl is the concentration of the acid
Explanation:
Based on the reaction, 1 mole of NaOH reacts per mole of HCl. That means the moles added of NaOH in the neutralization = Moles of HCl in the solution. With the moles and the volume in Liters we can find the molar concentration of HCl
<em>Moles NaOH = Moles HCl:</em>
25.0mL = 0.025L * (2.0moles / L) = 0.050moles HCl
<em>Molarity:</em>
0.050moles HCl / 0.0500L =
<h3>1.0M HCl is the concentration of the acid</h3>
Answer:
Explanation:
7 carbon atoms, 6 hydrogen atoms,9 NO2 atoms
7+6+9=22
2(22)=44
44 atoms
Answer:
The answer to your question is: yield = 56.27%
Explanation:
Data
CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br
18.54 ml 1-butanol 15.65 g of 1-bromobutane
% yield = ?
density = 0.81 g/ml
MM = 74 g 1- butanol
MM = 137 g 1-bromobutane
Process
Calculate mass of 1- butanol
density = mass/volume
mass = density x volume
mass = 0.81 x 18.54
mass = 15.02 g of 1-butanol
Theoretical yield
74 g of 1- butanol ----------------- 137 g of 1-bromobutane
15.02 g of 1- butanol ------------- x
x = (15.02 x 137) / 74
x = 27.81 g of 1-bromobutane
% yield = experimental yield / theoretical yield x 100
% yield = 15.65 / 27.81 x 100
% yield = 56.28
