In this solution liquid is solvent and the solid is solute. solid will be solve in liquid in normally but the increasing of tempreature make this solutuion speed higher because of preasure.
Answer:
The suitable answer to the given blank is
<u>Two reactions or pathways that share the substrate or products and result in no net gain of ATP. </u>
Explanation:
A futile cycle can be defined as those cycles which are involved in metabolism at cellular level to control or regulate biochemical pathways.
It is also known as substrate cycle and is when two metabolic pathways follow directions opposite to each other so that the overall effect is zero other than to dissipate energy.
Therefore, the result is zero net gain.
Answer:
Q = 3139.5 j
Explanation:
Given data:
Mass = 50 g
Initial temperature = 25°C
Final temperature = 95°C
Specific heat capacity = 0.897 j/g.°C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 95°C - 25°C
ΔT = 70°C
Q = m.c. ΔT
Q = 50 g× 0.897 J/g.°C ×70°C
Q = 3139.5 j
Answer:
n₁ = 2
Explanation:
The Rydberg formula is given by
1/λ = Rh x (1/n₁² - 1/n₂² ) where Rh = Rydberg´s constant , and n₁ and n₂ are the energy levels of the transitions involved.
The highest energy line in a series will happen when n2 tends to infinity, and then in
1/λ = Rh x (1/n₁² - 1/n₂² ) the term 1/n₂² in the limit is zero
1/λ = Rh x 1/n₁²
So lets solve for n₁² :
n₁² = Rh x λ = 109,737 cm⁻¹ x 365 x 10 ⁻⁷ cm = 4.00
n₁ = √4.00 = 2
This the value for the Balmer series.
<h3><u>Latin names:</u></h3>
Copper = Cuprum
Sulfur = Sulfurium
Tin = Stannum