If you cross a heterozygous organism with another heterozygous organism, pair two Tt and Tt. Below is a punnet square to show the result:
T t
T TT Tt
t Tt tt
This means that you have two heterozygous recessive and one homogenous dominant and one homogenous recessive.
Answer:
54 grams ammonium chloride and 40 grams sodium hydroxide
Explanation:
A buffer is a solution that contains either a weak acid and its salt or a weak base and its salt, the solution is resistant to changes in pH. This means that, a buffer is an aqueous solution of either a weak acid and its conjugate base or a weak base and its conjugate acid.
A Buffer is used to maintain a stable pH in a solution, buffers can neutralize small quantities of additional acid of base. For any buffer solution, there is always a working pH range and a set amount of acid or base that can be neutralized before the pH will change. The amount of acid or base that can be added to a buffer before changing its pH is called its buffer capacity.
A good buffer mixture is supposed to have about equal concentrations of its both components. It is a rule of thumb therefore, that a buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other component.
The implication of this is that the ammonium chloride and sodium hydroxide should be of approximately the same concentration. If the masses are dissolved as shown in the answer, then we will have 1molL-1 of each component of the buffer in accordance with the rule of thumb stated above.
The balanced reaction is
Na2O + H2O --> 2NaOH
If 2.24 moles of sodium oxide react, that means 4.48 moles of NaOH is formed as it is a 1 to 2 stoichiometric relationship.
Now we multiply by the molar mass to get grams.
4.48 moles NaOH * (39.997 grams/1 mole) = 179.2 grams
Your answer is 179. grams.
The light bulbs ave to be a part of the loop in order for it to be successful because it have to have wires to a circuit breaker so you can see.
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = ![\frac{ΔvapH}{R}([tex]\frac{1}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7B%CE%94vapH%7D%7BR%7D%28%5Btex%5D%5Cfrac%7B1%7D%7BT1%7D)
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
