Answer:
Half will be green and half will be striped
Explanation:
Given green skin (G) is dominant to striped skin (g).
Cucumber 1 : Heterozygous (Gg)
Cucumber 2 : Homozygous striped (gg)
When Gg X gg :
G g
g Gg gg
g Gg gg
Half of the offspring will have Gg genotype so they will be green. Other half will have gg genotype so they will be striped.
Is this multi choice or no?
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9
Answer:
There were 0.00735 moles Pb^2+ in the solution
Explanation:
Step 1: Data given
Volume of the KI solution = 73.5 mL = 0.0735 L
Molarity of the KI solution = 0.200 M
Step 2: The balanced equation
2KI + Pb2+ → PbI2 + 2K+
Step 3: Calculate moles KI
moles = Molarity * volume
moles KI = 0.200M * 0.0735L = 0.0147 moles KI
Ste p 4: Calculate moles Pb^2+
For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+
For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+
There were 0.00735 moles Pb^2+ in the solution
The correct option is; SUGAR DISSOLVING IN WARM WATER.
There are two types of change in chemistry; physical and chemical changes.
A physical change is a type of change in which no new substance is formed while a chemical change is a type of change in which a new substance is formed.
Dissolving sugar in warm water is a physical change because no new substance is formed and the sugar can easily be recovered from the water by evaporating the water.
