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PolarNik [594]
3 years ago
5

Name the process that changes matter into one or more new substances

Chemistry
1 answer:
Brums [2.3K]3 years ago
6 0
Chemical change or process
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How many calories are required to boil 75 grams of water not in sig figs
Norma-Jean [14]

Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius. (There are 1000 ml, each of which needs to have its temperature raised.)

3 0
2 years ago
How many moles of water are produced from 13.35 mol of oxygen?
olganol [36]

Answer:

The answer is 26.70

4 0
2 years ago
Part B Write the balanced chemical equation for the neutralization reaction that occurs when an aqueous solution of hydrobromic
Doss [256]

Answer:

HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

Explanation:

A neutralization reaction is a process in which an acid, aqeous HBr reacts completely with an appropriate amount of base, aqueous LiOH to produce salt, aqueous LiBr and water, liquid H2O only.

HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

Acid + base → Salt + Water.

During this reaction, the hydrogen ion, H+, from the HBr is neutralized by the hydroxide ion, OH-, from the LiOH to form the water molecule, H2O.

Thus, it is called a neutralization reaction.

4 0
2 years ago
A system of ideas that explains many related observations and is supported by a large body of evidence collected through scienti
wlad13 [49]
B. Theory
The hypothesis that stand the test of time (often tested and never rejected) is called theory. A theory is supported by a great dealcof evidence.
6 0
3 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
2 years ago
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