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Naddika [18.5K]
2 years ago
9

Is the cell wall factory part or worker? And please explain what the worker or factory part does. (Don’t tell me it’s on quizlet

bc I can’t find it)
Chemistry
1 answer:
dem82 [27]2 years ago
5 0
Cell wall is a worker because it allows water and sugar in and out (atp)
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You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale).
AVprozaik [17]

Answer:

Explanation:

It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.

7 0
2 years ago
A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
Tamiku [17]

Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow  Na_2SO_4(aq) + CO_2(g) + H_2O(l)

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

5 0
2 years ago
How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 years ago
Which describes any compound that has at least one element from group 17?
Contact [7]
The answer is HALIDE.
6 0
2 years ago
The concentration of ozone in a sample of air that has a partial pressure of O3 of 0.33 torr and a total pressure of air of 695
Goryan [66]

Answer:

0.047 %

Explanation:

Step 1: Given data

  • Partial pressure of ozone (pO₃): 0.33 torr
  • Total pressure of air (P): 695 torr

Step 2: Calculate the %v/v of ozone in the air

Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.

<em>%v/v = pO₃/P × 100%</em>

%v/v = 0.33 torr/695 torr × 100%

%v/v = 0.047 %

8 0
2 years ago
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