All categories

3 years ago

What is the function of the endoplasmic reticulum?

Chemistry

1 answer:

Semenov [28]3 years ago

8 0

Answer:

option d

Explanation:

it is a passage for the nucleus and the rest of the cell

You might be interested in

During which phase of the moon do neap tides occur?

Setler79 [48]

Quarter moon is the correct answer

4 0

3 years ago

Read 2 more answers

If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the

kolbaska11 [484]

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α + 4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

5 0

4 years ago

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed

KIM [24]

Answer:

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?

A) BrCH2CH2Br

B) CH3CH2CH2Br

C) CH3CHBr2

D) CH3CH2CH2CH3

E) BrCH2CH2CH2CH2Br

Explanation:

The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.

This is an example of free radical substitution.

The structure of ethane and its bromination is shown below:

Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).

Remaining all other products are possisble to form on free radical substitution of ethane.

8 0

3 years ago

Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1

jekas [21]

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

8 0

4 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago