Ask question

Ask question

All categories

3 years ago

10

Valerie is back in the lab to separate another blood sample, but her centrifuge has broken. Luckily, she remembers that red bloo

d cells have a large iron-based molecule at their center. What other method could she use to separate the blood cells?

2 answers:

Alex3 years ago

8 0

the answer is magnetic separation, not sedimentation separation

tangare [24]3 years ago

6 0

Answer:

magnetic seperation

Explanation:

You might be interested in

7) Muscles work to lift a bone in your arm by contracts and relaxes to<br><br><br>help please <br>

Andru [333]

Answer:

By contracting, muscles pull on bones and allow the body to move. When you want to bend your elbow, your biceps muscle contracts, and, at the same time, the triceps muscle relaxes. The biceps is the flexor, and the triceps is the extensor of your elbow joint.

Explanation:

yeah, hope it helps a little. :)

3 0

2 years ago

Aluminum chloride can be formed from its elements:

saul85 [17]

<u>Answer:</u> The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

6 0

3 years ago

Calculate the number of molecules in 100. g of O2 *

Sindrei [870]

I am pretty sure the answer is 6.25 molecules
See picture for explanation

6 0

3 years ago

What do El Niño and La Niña have in common? Check all that apply.

Yakvenalex [24]

Can you post a picture of that the checkings are

7 0

3 years ago

Read 2 more answers

Are plastic bags organic or inorganic?

Alborosie

Answer:

nilon is inorganic so they can be in the soil up to 5000 years

Explanation:

3 0

3 years ago

Read 2 more answers