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3 years ago

Muscles cells are called?

Chemistry

2 answers:

Marizza181 [45]3 years ago

8 0

muscles cells are called Myocyte .

evablogger [386]3 years ago

6 0

Answer:

They are called Myocyte.

Explanation:

A myocyte is said to be the type of cells that is found in muscle tissue. They are structurally long and also posses tubular cells that inhibits develop from myoblasts to form myogenesis process which are muscles.

Within a skeletal muscle fibre, there are many myocytes. arranged parallel to one another which gives the fibre a striated appearance. They are contractile parts of the muscle fibre.

Also, different forms of these myocytes has different qualities in our body muscles found in the cardiac, skeletal, and smooth muscle cells.

You might be interested in

Will a horse galloping in a field produce and average velocity of zero

motikmotik

Answer:

Explanation:

Depending on the mass of the horse and the speed, velocity will change.

5 0

3 years ago

Se construye una pila galvánica con una barra de cobre sumergida en una disolución 1M de cationes Fe+2 y una barra de plata sume

pashok25 [27]

Answer:

El potencial celular estándar, $E_{cell}$ is +0.46 V

Explanation:

Las reacciones de media célula son;

Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V

Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo

Por lo tanto, en el ánodo, tendremos

Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

$E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V$

El potencial celular estándar, $E_{cell}$ = +0.46 V

4 0

3 years ago

two samples of water, 10 grams and 50 grams are in sperate cups which of the following properties is different for the two boili

Trava [24]

Answer:

volume

Explanation:

The volume of the two samples of water will be different because volume is the amount of space occupied by a body. It is dependent on the amount of materials it contains.

The 50g sample will have a higher volume compared to the 10g sample.
The boiling point and density are intensive properties and do not depend on the amount of matter present.
Since both samples are from the same source, they will have the same color.

7 0

3 years ago

Read 2 more answers

The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p

zysi [14]

Answer : q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

$q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})$ (in terms of mass)

or, $q=\Delta H_{fusion}\times Moles$ (in terms of moles)

Now we have to calculate the value of q.

$q=6020J/mole\times 1Mole=6020J$

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of $\Delta e=0$

Now we have to calculate the value of w.

Formula used : $\Delta e=q+w$

where, q is heat required, w is work done and $\Delta e$ is internal energy.

Now put all the given values in above formula, we get

$0=6020J+w$

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 0

3 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.

See figure 1

I hope it helps!

4 0

3 years ago