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KATRIN_1 [288]
2 years ago
15

0.20 L of HNO3 is titrated to equivalence using 0.12 L of 0.2 M NaOH. What is the concentration of the HNO3?

Chemistry
1 answer:
AleksandrR [38]2 years ago
4 0

M_{A}V_{A}=M_{B}V_{B}\\(0.20)(M_{A})=(0.12)(0.2)\\M_{A}=\frac{(0.12)(0.2)}{(0.20)}=\boxed{0.12 \text{ M}}

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What is meant by saying that the latent heat of vaporization of water is 22. 5×10^5 j/kg.
andrey2020 [161]

It means that 22.5×10^5 J of heat is required to change 1 kg of water into steam.

Latent heat of vaporization is amount of energy required to change 1 gram (in this example 1 kilogram) of material from the liquid to the gaseous state at its boiling point.

Boiling point of the water is 100°C.

Joule (J) is the standard unit for energy (in this example heat).

Evaporization is phase change process in which the water changes from a liquid to a gas (water vapor). Fore example, solar radiation can be the source of energy for evaporation.

More about heat of vaporization: brainly.com/question/14679329

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5 0
2 years ago
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lawyer [7]

Answer:C

Explanation:

3 0
3 years ago
The O2 produced from the decomposition of the 1.0 L sample of H2O2 is collected in a previously evacuated 10.0 L flask at 300. K
Sergio [31]

The full question can be seen below:

2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}

The decomposition of H_{2}O_{aq} is represented by the equation above.

A student monitored the decomposition of a 1.0 L sample of H_{2}O_2_(_{aq}_) at a constant temperature of 300K and recorded the concentration of H_{2}O_2 as function of time. The results are given in the table below:

                                    Time (s)      H_{2}O_2

                                    0                 2.7

                                    200            2.1

                                    400            1.7

                                    600            1.4

The O_2_(_{g}_) produced from the decomposition of the 1.0 L sample of H_{2}O_2_(_{aq}_) is collected in a previously evacuated 10.0 L flask at 300 K. What is the approximate pressure in the flask after 400 s?

(For estimation purpose, assume that 1.0 mole of gas in 1.0 L exerts a pressure of 24 atm at 300 K).

Answer:

1.2 atm

Explanation:

Considering all assumptions as stated above;

                       2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}

Initial               2.7 mole               ---             ---

Change          -1.0                        ---             +\frac{1.0}{2}

Equilibrium     1.7 mole                ---             0.5 mole

To determine the concentration of O₂; we need to convert the moles to concentration for O₂ = \frac{0.5}{volume in the flask}

                                  = \frac{0.5 mol}{10.0 L}

                                  = 0.05 \frac{mol}{L}

Thus, based on the assumption that "1.0 mole of gas in 1.0 L exerts a pressure of 24 atm"

∴ 0.05\frac{mol}{L} will give rise to = 0.05 \frac{mol}{L} × 24

                                           = 1.2 atm

7 0
3 years ago
A molecule of antifreeze, ethylene glycol, has the formula c2h4(oh)2. how many atoms are there in one molecule of antifreeze?
kifflom [539]
Answer is: there are ten atoms in one molecule of antifreeze.
One molecule of ethylene glycol (C₂H₄(OH)₂) has two carbon atoms, six hydrogen atoms (4 + 2 · 1) and two oxygen atoms (2 · 1). So there are:
2 + 6 + 2 = 10 atoms.
Ethylene glycol (C₂H₄(OH)₂) is an odorless, sweet-tasting, colorless viscous liquid.
5 0
3 years ago
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mezya [45]
Since Chlorine is in excess, this is a limiting reagent problem.
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