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Anettt [7]
3 years ago
5

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.

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C. Greater exposure to radiation

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The equation shown represents cellular respiration. Glucose + X → Y + Water + Energy What do X and Y most likely represent? X re
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X represents oxygen and Y represents carbon dioxide.

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We observe a distant object in space and see that the spectral lines for hydrogen in the object's light appear at a shorter wave
MrMuchimi

Answer:

This tells us the radial velocity of the object and that the object is approaching or coming towards us.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
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