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4 years ago

Animal cells, including human cells, are considered:

1 answer:

3 0

Answer:Animal cells are typical of the eukaryotic cell, enclosed by a plasma membrane and containing a membrane-bound nucleus and organelles. Unlike the eukaryotic cells of plants and fungi, animal cells do not have a cell wall.

Explanation:

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2 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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3 years ago

What is the molarity of the Ca(OH)2 solution if 32.00 mL of Ca(OH)2 requires 16.08 mL of a 2.303 M solution for complete titrati

amm1812

Answer:

For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb

Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L

So now we can plug into the equation and solve:

1.338 mol/L * 0.01875 L= x mol/L * 0.02473 L

This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.

First step: setup the stoichiometry problem:

3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl

Second step: Molarity equation

Molarity= moles HCl/ L M= mol HCl/0.04027 L

For the third problem, you will just use the same equation as the first: MaVa= MbVb

Ma=0.57 M Va= x L Mb= 0.875 M Vb= 23.83 mL= 0.02383 L

0.57 M * x L= 0.875 M * 0.02383 L

With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.

0.75 M= x mol/ 0.0227 L mol NaOH = 0.75 M *0.0227 L mol NaOH= 0.017025

So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles

Molar Mass= 3.6 g/ 0.017025 mol

And with that you will find the molar mass of HX, and even determine what X is.

Explanation:

hope I helped

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3 years ago