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Ratling [72]
3 years ago
11

Is a burning log an exothermic or endothermic event if the log is the system?

Chemistry
2 answers:
ziro4ka [17]3 years ago
7 0
Yes it is a exothermic reaction.
Anestetic [448]3 years ago
4 0
Exothermic because its burning and it releases energy.
You might be interested in
Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.
Sophie [7]

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

y=-3.5 x + 25\\\\y=-0.52 x + 2

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

y=-3.5 (7.72) + 25\\\\y = 1.84

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

Learn more:

  • brainly.com/question/22818252
  • brainly.com/question/9680530

7 0
3 years ago
The picture below shows a warm air mass caught between two cooler air masses. What is this type of front called? Plz, I really n
pentagon [3]

An occluded front forms when a warm air mass is caught between two cooler air masses.

6 0
3 years ago
A) release of toxic gas
faust18 [17]
In my opinion the answer is D
5 0
3 years ago
Read 2 more answers
What are the characteristics of an experiment? What are key parts of an experiment?
Alecsey [184]

Answer:

independent variable - the thing you change

dependent variable - the thing you measure

control variables - the things you keep the same

6 0
3 years ago
• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

6 0
3 years ago
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