Calculate [H 3 O+] of a solution with the ph of 3.45

Alguien sabe algo de introducción ala fisica ?

pshichka [43]

Answer: Un ala es un tipo de aleta que produce elevación, mientras se mueve a través del aire o algún otro fluido. Como tal o puede calcularse a partir de la distribución de velocidad aérea utilizando principios físicos básicos

Explanation:

3 0

3 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

4 years ago

Read 2 more answers

Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil

quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

$Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}$

7 0

3 years ago

Which expression represents a reaction rate?a. time/massc. energy/timeb. number/timed. time/energy

Kitty [74]

The right answer for the question that is being asked and shown above is that: "b. number/timed." Reaction Rate refers to the speed of reaction for a reactant or product in a particular reaction is intuitively defined as how fast or slow are action takes place.

4 0

3 years ago

Read 2 more answers

Question:

Dahasolnce [82]

If you can have more than 1 answer then the dependent variable is graphed on the Y-axis and and is usually used for comparison since they are usually the experiment results. If you can only have one answer then the dependent variable is graphed on the y-axis.

8 0

3 years ago