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3 years ago

13

Muscles convert chemical energy into

1 answer:

Sergeu [11.5K]3 years ago

8 0

Muscles convert chemical energy into mechanical energy

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Give the reason for the following; 1. Iodimetric titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly

Agata [3.3K]

Answer:

Explanation:

blach bal

5 0

3 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

4 0

3 years ago

On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i

ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (12 + 273.15) K = 285.15 K

T₂ = (32 + 273.15) K = 305.15 K

Using above equation as:

$\frac{1.5}{285.15}=\frac{V_2}{305.15}$

$V_2=\frac{1.5\cdot \:305.15}{285.15}$

New volume = 1.6 L

8 0

3 years ago

A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

ioda

In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = $\frac{PV}{nT}$ = constant
$\frac{P1 V1}{n T}$ = $\frac{P2 V2}{n T}$
==> P1 * V1 = P2 * V2
V2 = $\frac{P1 V1}{P2}$ = $\frac{445 0.016}{1225}$
V2 = 0.00581 m3 = 5.81 L

7 0

4 years ago

What means that the "p" orbital has 3 orientations?

Vlada [557]

Answer:

The p orbital is a lobed region describing where an electron can be found, within a certain degree of probability. All p orbitals have l = 1, with three possible values for m (-1, 0, +1).

Explanation:

3 0

3 years ago