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Genrish500 [490]
3 years ago
11

Cual es la masa atomica del hidrogeno

Chemistry
1 answer:
Fittoniya [83]3 years ago
4 0

El hidrógeno es el elemento químico de número atómico 1, representado por el símbolo H. Con una masa atómica de 1.00784 u  ​ es el más ligero de la tabla periódica de los elementos. Por lo general, se presenta en su forma molecular, formando el gas diatómico H₂ en condiciones normales.

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Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit
Anastasy [175]

The dissociation equation will be

                         NH4OH   --->        NH4+                   + OH-

Initial                 0.006                        0                          0

Change         -0.006 X 0.053        +0.006 X 0.053      -0.006 X 0.053

Equlibrium     0.006 -0.006 X 0.053      0.006 X 0.053    0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0
3 years ago
Which of the following are compounds (select all that apply)
olga2289 [7]

Answer:

KNO2, KBr

Explanation:

Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.

3 0
2 years ago
What are some other examples of joints in your body that work like a hinge?
Dmitrij [34]
Knees and elbows are two of the most common ones.
8 0
4 years ago
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In nature amino acids exist in what type of conformation?
liberstina [14]

Answer:

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Explanation:

7 0
3 years ago
An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the
sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

P_i = 100 N·m

T_i = 327 K

P_f = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln(P_f/P_i) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈<em> -0.52571</em> kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln(P_f/P_i) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = <em>-7.5 kJ</em>

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = <em>-5.00574 kJ</em>

6 0
3 years ago
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