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love history [14]
3 years ago
9

Fermeroben heres the messages COUGH

Chemistry
1 answer:
lord [1]3 years ago
6 0

I completely understand your gripe. I also believe that you are the better option. It will only take time until she is yours, despite your parent's views. Keep it up

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Complex chemical catalysts produced by living cells are called enzymes. true false
SIZIF [17.4K]
True 
Enzymes help the body in digestion and other bodily functions which involve chemicals.
7 0
3 years ago
Read 2 more answers
How many molecules of CaCi2 are equivalent to 75.9 grams
horsena [70]
<em>M CaCl₂: 40+(35,5×2) = 111 g/mol</em>


6,02·10²³ molecules ---------- 111g
X molecules --------------------- 75,9g
X = (75,9×<span>6,02·10²³)/111
X = <u>4,116</u></span><span><u>·10²³</u> molecules of CaCl</span>₂

:)
8 0
3 years ago
In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form
iris [78.8K]

Answer:

I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.

Explanation:

<em>I verify this is correct. </em>

6 0
3 years ago
true or false in a chemical reaction the same number of each type of atom will always be present before and after the reaction t
iren [92.7K]

Answer:

True

Explanation:

True because of the law of conservation of mass, the same same amount of atoms will be on both sides of the reaction.

6 0
2 years ago
Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f
Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



4 0
3 years ago
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