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Alex777 [14]
2 years ago
7

Why do heat (q) and work (w) have positive values when entering a system and negative values when leaving?

Chemistry
1 answer:
Papessa [141]2 years ago
3 0

Energy transferred into the system is wonderful due to the fact the system ends up with greater strength while energy transferred out from the machine is negative, because the device finally ends up with much less energy.

In defining work, we awareness at the outcomes that the system (e.g. an engine) has on its surroundings. for this reason we outline work as being fine when the gadget does work at the environment (strength leaves the machine). If work is completed at the gadget (energy added to the machine), the work is bad.

The classical sign convention states that warmth switch into a device and work produced with the aid of it are tremendous, while heat transfer faraway from a device and work produced on it is negative.

Q is wonderful if warmness is added into the device from the surroundings. The take a look at tube would feel cooler if the response had been being finished in one. If the surroundings is tormented by the machine, work is poor. This indicates that energy is escaping the system.

Learn more about work  brainly.com/question/18094932

#SPJ4

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You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
34.2 gram of sucrose is dissolved in 180 gram water. Calculate the number of hydrogen and oxygen atoms present in the solution.
Naddika [18.5K]

Explanation:

34.2g of C12H22O11 is dissolved in 180g of H20.

Molar mass of sucrose = 342g/mol

Moles of sucrose = 342 / 34.2 = 10 mol.

Molar mass of water = 18g/mol

Moles of water = 180 / 18 = 10 mol.

For hydrogen atoms, there are 22 * 10 in sucrose and 2 * 10 in water, which gives a total of 240.

For oxygen atoms, there are 11 * 10 in sucrose and 1 * 10 in water, which gives a total of 120.

8 0
3 years ago
Help me guys plz with both of them
Nimfa-mama [501]
First one is a
the
second one is b
5 0
3 years ago
Read 2 more answers
Which of the following is a homogeneous mixture?A) Molten ironB) Stainless steelC) Trail mixD) Water
ehidna [41]

Answer:

B) Stainless steel

Explanation:

Homogeneous mixture  -

It refers to the composition of various substance , which leads to a mixture of complete uniform composition , i.e. , the ratio of each and every component of the mixture is constant all over the mixture , is referred to as a homogeneous mixture.

For example ,

In the solution of sugar and water , where 1 tablespoon of sugar is dissolved in 100 mL of water ,

The ratio of sugar and water remains constant all over the mixture.

Hence, from the given options the homogeneous mixture is B) Stainless steel .

5 0
3 years ago
Consider the reaction:
Fofino [41]

The mass in grams of NH₃ produced from the reaction is 3.4 g

<h3>Balanced equation</h3>

We'll begin by writing the balanced equation for the reaction. This illustrated below:

N₂ + 3H₂ -> 2NH₃

From the balanced equation above,

1 dm³ of N₂ reacted to produced 2 dm³ NH₃

<h3>How to determine the volume of NH₃ produced</h3>

From the balanced equation above,

1 dm³ of N₂ reacted to produced 2 dm³ NH₃

Therefore,

2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃

<h3>How to determine the mass of NH₃ produced</h3>

We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:

22.4 dm³ = 1 mole NH₃

Therefore,

4.48 dm³ = 4.48 / 22.4

4.48 dm³ = 0.2 mole of NH₃

Finally, we shall determine the mass of NH₃ as follow:

  • Molar mass of NH₃ = 17 g/mol
  • Mole of NH₃ = 0.2 mole
  • Mass of NH₃ =?

Mass = mole × molar mass

Mass of NH₃ = 0.2 × 17

Mass of NH₃ = 3.4 g

Learn more about stoichiometry:

brainly.com/question/13196642

#SPJ1

4 0
1 year ago
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