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4 years ago

How would these cells appear under a microscope at a higher magnification.

Chemistry

1 answer:

gulaghasi [49]4 years ago

5 0

they would be more detailed? what type of cells though?

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Intermolecular forces dipole differences london dispersion

loris [4]

33233728793278237876548742787874578378572098-2932-=93788784787489

8 0

3 years ago

How man grams of cl2 are consumed to produce 12.0 g of KCl

Korvikt [17]

Answer:

5.71 g

Explanation:

Step 1: Write the balanced equation

2 K + Cl₂ ⇒ 2 KCl

Step 2: Calculate the moles corresponding to 12.0 g of KCl

The molar mass of KCl is 74.55 g/mol.

12.0 g × 1 mol/74.55 g = 0.161 mol

Step 3: Calculate the moles of Cl₂ needed to produce 0.161 moles of KCl

The molar ratio of Cl₂ to KCl is 1:2. The moles of Cl₂ needed are 1/2 × 0.161 mol = 0.0805 mol

Step 4: Calculate the mass corresponding to 0.0805 moles of Cl₂

The molar mass of Cl₂ is 70.91 g/mol.

0.0805 mol × 70.91 g/mol = 5.71 g

8 0

3 years ago

Use the table on the right to calculate each required quantity.

k0ka [10]

a. 35.8 KJ

b. 871 g

c. Fe

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3 0

3 years ago

Q #13 How many moles of MgCl2 are there in 350. g of compound?

Vaselesa [24]

<u>Answer:</u>

3.67 moles

<u>Step-by-step explanation:</u>

We need to find out the number of $MgCl_2$ moles present in 350 grams of a compound.

Molar mass of $Mg$ = 24.305

Molar mass of $Cl_2$ = 35.453

So, one mole of $MgCl_2$ = 24.305 + (35.453 * 2) = 95.211g

1 Mole in 1 molecule of $MgCl_2$ = $\frac{1}{95.211} = 0.0105$

Therefore, number of moles in 350 grams of compound = 0.0105 * 350

= 3.67 moles

8 0

4 years ago

Tyrosine absorbs maximally at 290 nm. A 0.2162 g sample containing some amount of Tyrosine was placed into a 250.0 mL volumetric

mojhsa [17]

Answer:

B. 0.069 %

Explanation:

It should be initially noted in this answer in particular that, the amino acids tryptophan and tyrosine have a very precise 280 nm absorption rate, which allows a direct A280 size of protein concentration. The 280 nm UV absorbance rate is regularly utilized to approximate protein concentration in laboratories due to its simplistic nature, its affordability and also the ease of usage.

kindly check the attached image below for the solution to the above question.

7 0

3 years ago