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3 years ago

6

Calcium ions are removed from the sarcoplasm of the cardiac cell by the action of the sarcoplasmic reticulum (SR) Ca2+-ATPase pu

mp and which other mechanism?

1 answer:

Ivan3 years ago

7 0

Answer:

Sodium-calcium exchanger on the sarcolemma

Explanation:

The sodium-calcium exchanger which is an antiporter membrane protein, removes calcium from cells. The energy of the electrochemical gradient of sodium is used by allowing it to flow down its gradient while moving across the plasma membrane

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

!!!!HURRY!!!! What reaction is endothermic?

Alex

Answer:

The reaction in which heat is absorbed from the surrounding is called endothermic reaction.

so

I think it's answer is 2nd option.

4 0

3 years ago

Which of the following is NOT a characteristic of stars? A) Size B) temperature C) texture D) color

Lynna [10]

Answer:

C

Explanation:

Characteristics used to classify stars include color, temperature, size, composition, and brightness.

7 0

3 years ago

Complete the table to summarize the properties of the different subatomic particles. Type in your answers.

Aloiza [94]

<h2>Complete the table to summarize the properties of the different subatomic particles. </h2>

Explanation:

Atom

It is a smallest particle which cant exist independently.

According To Dalton, atom was indivisible but later on, it was proved that atom can be subdivided into sub atomic particles called electron, proton & neutron.

These subatomic particles have marked properties .

Proton

It was discovered by E.Goldstein .
It is positively charged particle
It is present in nucleus .
Its mass is equal to 1.6726219 × 10⁻²⁷ kilograms

Neutron

It was discovered by E.chadwick .
It is neutral
It is present inside the nucleus .
It's mass is equal to 1.674927471×10⁻²⁷ kg

Electron

It was discovered by J.J Thomson .
It has negative charge .
It's mass is equal to 9.10938356 × 10⁻³¹ kilograms
It is present outside the nucleus in shells .

4 0

2 years ago

Read 2 more answers

A chemistry teacher is able to grade chemistry labs at a rate of 5 labs for every 3 minutes. They need to grade 165 labs how man

PIT_PIT [208]

Answer:

1.65hr

Explanation:

Given parameters:

Number of labs = 5 labs

Time taken = 3 minutes

Unknown:

Time taken in hours to grade 165 labs = ?

Solution:

Let us find the rate of the teacher;

Rate = $\frac{number of labs}{time}$

Insert the parameters and solve;

Rate = $\frac{5}{3}$ = 1.67labs/min

Now;

Time to grade 165 labs = $\frac{number of labs}{rate}$

Time to grade 165 labs = $\frac{165}{1.67}$ = 98.8min

Since;

60min = 1hr

98.8min = $\frac{98.8}{60}$ = 1.65hr

3 0

3 years ago