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aksik [14]
3 years ago
10

Explain where you would find the three different types of muscles (cardiac, smooth and skeletal) and how these different types o

f muscles are used in the body.
Chemistry
1 answer:
yawa3891 [41]3 years ago
4 0

Explanation: Skeletal muscle, attached to bones, is responsible for skeletal movements. The peripheral portion of the central nervous system (CNS) controls the skeletal muscles. Thus, these muscles are under conscious, or voluntary, control. The basic unit is the muscle fiber with many nuclei. These muscle fibers are striated (having transverse streaks) and each acts independently of neighboring muscle fibers.

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g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i
kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is  \alpha_{Y^{-4}} =0.81

So the Conditional Formation constant= K_f =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL  

V e= 25.00 mL  

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.  

[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M

                                                            {K^'}_f

                       Ca²⁺      +      Y⁴          ⇄     CaY²⁻

Initial                 0                  0                      0.001667

change             +x                  +x                     -x

equilibrium        x                    x                    0.001667 - x

{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\

x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M  

pca = —log(2 15x101= 6.7

3 0
4 years ago
Please answer this as soon as possible Thank You.
sashaice [31]

Answer:

jchb3prhoucheoucheoucheouchoudehocuehcdouhd

Explanation:

5 0
3 years ago
In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53 × x10−
sergiy2304 [10]

the force between the electron and the proton.
a) Use F = k * q1 * q2 / d² 
where k = 8.99e9 N·m²/C² 
and q1 = -1.602e-19 C (electron) 
and q2 = 1.602e-19 C (proton) 
and d = distance between point charges = 0.53e-10 m 
The negative result indicates "attraction". 

the radial acceleration of the electron. 
b) Here, just use F = ma 
where F was found above, and 
m = mass of electron = 9.11e-31kg, if memory serves 
a = radial acceleration 

the speed of the electron. 
c) Now use a = v² / r 
where a was found above 
and r was given 

<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v 
where v was found above 
and r was given 
3 0
3 years ago
The most stable conformation of trans-1-tert-butyl-2-methylcyclohexane is the one in which:
Margaret [11]

The most stable conformation of trans-1-tert-butyl-2-methylcyclohexane is the one in which both the tert-butyl group and the methyl group are located near the equatorial position. Trans conformation are usually stable at the equatorial position to avoid the bulkyl group ( the tert-butyl group and methyl group) be together to reduce the so called Steric hindrance.

4 0
3 years ago
The solution you identified in question (1) acts as a buffer due to reactions that occur within the solution when an acid or a b
weeeeeb [17]

Answer:

The answer is "\bold{CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)}"

Explanation:

When HCI is added in the chemical equation it reacts with sodium acetate so, it will give the following chemical equation:

CH_3COONa\ (aq) + HCl\ (aq)\longrightarrow CH_3COOH\ (aq) + NaCl\ (aq)\\\\

 In this, the CH_3COOH is a weak acid so, it not completely dissociated.

CH_3COONa \ (aq) \ \ and \ NaCl were strong electrolytes they are completely dissociated.

The HCl is a strong acid so, it is completely dissociated So, the net ionic equation is:

CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)

8 0
3 years ago
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