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Evgen [1.6K]
2 years ago
9

Exchanging ideas about science is called ______?

Chemistry
1 answer:
sesenic [268]2 years ago
5 0
Communication.

When people exchange ideas about science, they are communicating with one another to both share and obtain knowledge. Collaboration is a large part of this communication, as the exchange of these ideas often leads to new discoveries and inventions.
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Which colour would you expect ekasilicon to be
LiRa [457]

Answer:

it would ether be gray to black or silver to gray and black

Explanation:

because of my calculation ekasillicon would be that color because of the eyes and because of your color seeing .

7 0
2 years ago
Does Kinetic energy depend only on an object’s speed and velocity?
BaLLatris [955]

yes         .. . . . .. . . .. ....... . ..dlnv3r;'mw,c    kc;oqc,

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4 0
2 years ago
Read 2 more answers
A potential energy diagram is shown.
Vesna [10]

Answer:

25kJ

Explanation:

Given the initial energy to be 30kJ

The energy change from the initial energy to the peak energy = (65-30) kJ

= 35kJ

since the second energy change was a drop in energy it is regarded negative

= (55-65)

= -10kJ

Therefore total energy change

= (35-10)kJ

= 25kJ

5 0
2 years ago
How to balance Mg3n2+H20>>Mg(oh)2+Nh3?
alexandr1967 [171]
Mg₃N₂ + 6H₂O  =  3Mg(OH)₂ + 2NH₃
4 0
3 years ago
If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
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