Answer:
Dominant genes are always represented by a capital letter which in this case is the letter B whereas recessive genes will be represented by a lower case letter like b.
You only need one copy of a dominant gene or allele to inherit it. A dominant allele is always expressed even if only one copy is present.
But you need two copies of genes or alleles to inherit the recessive gene. There also can't be a dominant allele of the gene if you are to inherit this.
Allele - Different forms of a gene
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x
Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5
The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.
K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>
Answer:
It's C
Explanation:
Most garnet found near Earth's surface forms when a sedimentary rock with a high aluminum content, such as shale, is subjected to heat and pressure intense enough to produce schist or gneiss.
Answer:
5.37 L
Explanation:
To solve this problem we need to use the PV=nRT equation.
First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:
- P = 785 mmHg ⇒ 785/760 = 1.03 atm
- T = 18 °C ⇒ 18 + 273.16 = 291.16 K
1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K
We <u>solve for n</u>:
Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).
- 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K
And we <u>solve for V</u>:
