Answer:
animal cells dont have chloroplast
Explanation:
chloroplast is used to produce chlorophyll, only something found in plant cells
Ranking of the ions in each set in order of decreasing size -
Cs+ > Ba²⁺ > Sr²⁺
Cs+ has the biggest size, because its more downward (compared to Sr2+) and more to the left (compared) ot Ba2+.
Sr2+ has the smallest size because it's more upwords (compared to Cs+ and Ba2+).
For isoelectronic ions, the greater the nuclear charge , the greater is the attraction for electrons and smaller is ionic radius. In other words, size of the isoelectronic ions decreases with increase in atomic number.
Cl3+ has the smallest ion among all.
To learn more about ions from given link
brainly.com/question/14658491
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Answer:
c. 0.2 M HNO₃ and 0.4 M NaF
.
Explanation:
A buffer is defined as the mixture of a weak acid with its conjugate base or a weak base with its conjugate acid.
A weak acid or weak base are defined as an acid or base that partially dissociates in aqueous solution. in contrast, a strong acid or base are acids or bases that is dissociated completely in water.
Thus:
a. 0,2M HNO₃ and 0.4 M NaNO₃. This is a mixture of a strong acid with its conjugate base. <em>IS NOT </em>a buffer.
b. 0.2 M HNO₃ and 0.4 M HF
. This is a mixture of two strong acids. <em>IS NOT </em>a buffer.
c. 0.2 M HNO₃ and 0.4 M NaF
. NaF is the conjugate base of a weak acid as HF is.
The reaction of HNO₃ with NaF is:
HNO₃ + NaF → HF + NaNO₃
That means that in solution you will have a weak acid (HF) with its conjugate base (NaF). Thus, this mixture <em>IS </em>a buffer.
d. 0.2 M HNO₃ and 0.4 M NaOH. This is the mixture of a strong acid with a strong base, thus, this <em>IS NOT </em>a buffer.
I hope it helps!
We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal
