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WITCHER [35]
2 years ago
6

How are each of these seeds most likely spread from place to place

Chemistry
1 answer:
NISA [10]2 years ago
6 0
Try this website 
http://sciencelearn.org.nz/Science-Stories/Seeds-Stems-and-Spores/Seed-dispersal
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A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the
Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}

R is Gas constant having value = 0.008314 kJ / K mol  

Given temperature, T = 293 K

Given, \Delta{G^0}=15\ kJ/mol

So,  Applying in the equation as:-

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

Thus,

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}

\frac{[His]}{[His+]}=0.00211

Also, considering:-

pH=pKa+log\frac{[His]}{[His+]}

Given that:- pH = 7.0

So, 7.0=pKa+log0.00211

<u>pKa of the histidine = 9.67</u>

8 0
2 years ago
The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12
Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu

7 0
2 years ago
What properties are those that describe what happens when a substance reacts with another substance? catalyst properties chemica
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............ chemical properties.........
5 0
3 years ago
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What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3
Allisa [31]

O.N. of Na = +1

O.N. of O = -2

Let, O.N. of Tin = x

1*2 + x + -2*2 = 0

2+x-4 = 0

x-2 = 0

x = 2

SO OPTION C IS YOUR ANSWER......

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3 years ago
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In a heating curve, when is the temperature constant? need answers!
Varvara68 [4.7K]
During a phase change
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