Answer:
The correct answer is - they can create genetic diversity as well and reproduce without mate when necessary.
Explanation:
Sexual reproduction provides an organism with genetic diversity and variation by the process and it required two mates and a longer time to pollination and fertilization and is used in normal conditions.
In case of a threat, such organisms use asexual reproduction to increase their number as in asexual reproduction no need of mate, an organism can grow and increase on its own it provides to not to exitinct.
Answer:
A. 266g/mol
Explanation:
A colligative property of matter is freezing point depression. The formula is:
ΔT = i×Kf×m <em>(1)</em>
Where:
ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g
Replacing in (1):
0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg
<em>0,0614 = mol of solute</em>.
As molar mass is defined as grams per mole of substance and the compound weights 16,0g:
16,0g / 0,0614 mol = 261 g/mol ≈ <em>A. 266g/mol</em>
I hope it helps!
Its B i believe, Storms and currents can bring in sediments from other places
Answer:
Filtering <span>is best laboratory technique to separate a solid from a liquid to recover the liquid.
Explanation:
A solid particles present in liquids can be separated from liquid by utilizing a membrane having pores large enough to allow the liquid molecules to pass through and small enough to stop the solid particles from crossing.
Example:
Tea Filters are used to separate Tea from the grounds.
Kidney is an excellent example of Biological Filter.
</span><span>Whatman Cellulose Filter Paper used in Chemistry Labs.</span>
Answer: V2= 0.224L
Explanation:
P1=3.55atm, V2= 7.35L, T1= 290, P2= 123atm, V2=?, T2= 307K
Applying general gas equation
P1V1/T1= P2V2/T2
Substitute and Simplify
3.55×7.35/290 = 123×V2/307
V2= 0.224L
