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Vlad [161]
3 years ago
8

If a drug caused a sudden release of large amounts of calcium in skeletal muscle, from which structure would the calcium come?

Chemistry
2 answers:
lawyer [7]3 years ago
7 0
<span>Sarcoplasmic reticulum</span>
ser-zykov [4K]3 years ago
4 0

Answer:

The correct answer will be option-sarcoplasmic reticulum.

Explanation:

The sarcoplasmic reticulum is the membrane bound endoplasmic reticulum of the muscle cell which stores a large amount of Ca⁺² ions.

The reticulum contains Calsequestrin- a calcium binding protein which binds around 50 Ca⁺² ions to itself and helps store more calcium. These calcium ions are released when the signals through T-tubules are carried to the reticulum which leads to the opening of the calcium channels.

Thus, sarcoplasmic reticulum is the correct answer.

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What interactions are responsible for maintaining quaternary protein structure? Select all that apply.
sp2606 [1]
The ones that apply are A and C
4 0
3 years ago
Atoms share one or more pairs of electrons.<br><br> Options:<br> Ionic Bond or Covalent Bond?
garri49 [273]

Answer:

covalent bonds

Explanation:

ionic transfer of e^- ions formed (charges)

ionic=non-metal+ metal

ex: F+Ca

covalent sharing e^- no true charges

covalent= non-metal+ non-metal

ex: F+P

( my notes)

5 0
2 years ago
Which of the following is a radioisotope used to date rock formations older than 50,000 years old?
Eddi Din [679]
<span>The radioisotope used to date rock formations 50 000 years ago is Uranium. This radioactive uranium isotope having a mass number of 235 and its symbol is U, its atomic number is 92, and the mass number is 238, comprising 0.715 percent of natural uranium. When bombarded with neutrons it undergoes fission with the release of energy.</span>
7 0
3 years ago
How many grams of phosphorus are in 500.0 grams of calcium phosphide? (i need the work also)
kvv77 [185]

Answer:

\boxed{\text{170.0 g P}} 

Explanation:

The formula of calcium nitride is Ca₃P₂.

The masses of each element are:

\begin{array}{lrcr}\text{3Ca:} & 3 \times 40.08&=& \text{120.24 u}\\\text{2P:} & 2\times 30.97&=& \text{61.94 u}\\& \text{TOTAL} & = & \text{182.18 u}\\\end{array}

So, there are 61.94 g of P in 182.18 g of Ca₃P₂.

In 500 g of Ca₃P₂:

\text{Mass of P} = \text{500.0 g Ca$_{3}$P$_{2}$} \times \dfrac{\text{61.94 g P}}{\text{182.18 g Ca$_{3}$P$_{2}$}} = \text{170.0 g P}

There are \boxed{\textbf{170.0 g P}} in 500.0 g of Ca₃P₂.

3 0
3 years ago
Answer question number 6
Romashka [77]
It will be the first one
4 0
3 years ago
Read 2 more answers
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