Answer:
Answer) The first box represent the budding type of asexual reproduction . In budding due to the cell divison at the particular place in the organism , new individual will produced.
Second picture represent the budding in hydra. Budding is also a asexual reproduction.
Third type of asexual reproduction is found in plants it is vegetative reproduction.
Fourth is Binary fission . This is also a type of asexual reproduction, in which nucleus dicied into two new daughter cell via mitosis.Explanation:
819 ml is the volume the gas will occupy at 30.0°C.
Explanation:
Data given:
initial volume of the gas V1 = 590 ml
initial temperature of the gas T1 = -55 degrees OR 218.15
final volume of the gas V2 = ?
final temperature of the gas T2 = 30 degrees OR 303.15
Charles' Law equation is used to calculate the volume of gas at 30 degrees from the data given in the question.
= 
V2 = 
V2 = 
V2 = 819 ml
The final volume of the gas would be 819 ml.
Missing question: what is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00L gives 0.169 M NaOH?
Answer is: density is 1.52 g/mL.
c₁(NaOH) = ?; molarity of concentrated sodium hydroxide.
V₁(NaOH) = 16.7 mL; volume of concentrated sodium hydroxide.
c₂(NaOH) = 0.169 M; molarity of diluted sodium hydroxide.
V₂(NaOH) = 2.00 L · 1000 mL/L = 2000 mL; volume of diluted sodium hydroxide.
Use equation: c₁V₁ = c₂V₂.
c₁ = c₂V₂ / V₁.
c₁ = 0.169 M · 2000 mL / 16.7 mL.
c₁(NaOH) = 20.23 M.
m(NaOH) = 20.23 mol · 40 g/ml.
m(NaOH) = 809.53 g.
The mass fraction is the ratio of one substance (in this example sodium hydroxide) with mass to the mass of the total mixture (solution).
Make proportion: m(NaOH) : m(solution) = 53.4 g : 100 g.
m(solution) = 1516 g in one liter of solution.
d(solution) = 1516 g/L = 1.52 g/mL.
Answer:
The answer is C (or 3 in this case)
Explanation:
I took the test and got it correct and the explanation is that the watershed will flow into more water which spreads that pollution and as it flows could pick up more pollution.
<u>Answer: </u>The volume occupied by 
at STP is 0.56 L.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
We are given:
Given mass of oxygen = 0.80 g
Molar mass of oxygen = 32 g/mol
Putting values in above equation, we get:
<u>At STP:</u>
1 mole of a gas occupies 22.4 L of volume.
So, 0.025 moles of oxygen gas will occupy = 
of volume.
Hence, the volume occupied by at STP is 0.56 L.
