Answer:
V H2O = 170.270 mL
Explanation:
- QH2O ( heat gained) = Qcoffe ( heat ceded)
⇒ Q = m<em>C</em>ΔT
∴ m: mass (g)
∴ <em>C</em>:<em> </em>specific heat
assuming:
- δ H2O = δ Coffe = 1.00 g/mL
- <em>C</em> H2O = <em>C</em> coffe = 4.186 J/°C.g....from literature
⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)
∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe
⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J
⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)
⇒ (26371.8 J)/(154.882 J/g) = m H2O
⇒ m H2O = 170.270 g
⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL
Answer:
a. What makes this an oxidation-reduction reaction? (1 point)
Loss of electron by the aluminum and the gain of electron by the silver
b. Write the half-reactions showing the oxidation and reduction reactions. Identify which is the oxidation reaction and which is the reduction reaction. (3 points)
oxidation half reaction: Al → Al3+ + 3e-
Reduction half reaction: Ag+ + e- → Ag
c. What is oxidized in the reaction? What is reduced? (2 points)
Aluminum is oxidized and silver is reduced
d. In this simple electrochemical cell, what functions as the anode? What is the cathode? (3 points)
In a simple electrochemical cell the electrode where oxidation takes place is the anode. And the electrode where reduction reaction happens is the cathode.
e. Is this a galvanic cell or electrolytic cell? Explain your answer. (2 points)
An electrolytic cell because the reaction converts electrical energy into chemical energy
Not sure about e and f!
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
Answer: Extracellular [Ca2+]
Explanation:
The sensitivity and density of the alpha receptors serve to <em>enhance the response to the release of</em> <em>norepinephrine (NE)</em> . However, they do not exert a strong influence as the concentration of calcium ions on the amount of <em>norepinephrine (NE)</em> released by sympathic nerve terminals.
The release of neurotransmitters depends more on either an external or internal stimulus.This results in an action potential which on reaching a nerve terminal, results in the opening of Ca²⁺ channels in the neuronal membrane. Because the extracellular concentration of Ca²⁺ is greater than the intracellular Ca²⁺ concentration, Ca²⁺ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing <em>norepinephrine (NE)</em> to fuse with the plasma membrane and release <em>norepinephrine (NE)</em> into the synapse. The higher the action potential, the higher the Ca²⁺ flow into the terminals resulting in higher amount of <em>norepinephrine (NE)</em> into the synapse, and vice versa.
Catechol-O-methyltransferase (COMT) is one of several enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. It serves a regulatory purpose to lower the concentration of norepinephrine upon its release from nerve terminals.
Answer:
Static electricity is the result of an imbalance between negative and positive charges in an object. ... When two materials are in contact, electrons may move from one material to the other, which leaves an excess of positive charge on one material, and an equal negative charge on the other.
