Chem question!! please help!!!

If starch, glucose and NaCl are soaked in the dialysis tubing , will there be a difference between the beginning and after 15min

spin [16.1K]

Yes. If you put solutions containing the same concentration table salt, glucose and starch in different bags and place the bags in water (i.e., 0M) solution, then the bag with salt will contain more water after 15 minutes than the bag with glucose, which will contain more than the bag with starch.

6 0

3 years ago

What does applied chemistry have to do with making ice cream?

Lapatulllka [165]

Well you have to mix chemicals which caused a chemical reaction. When you mix salt and ice the freezing point of liquid changes. the ice gets colder

6 0

3 years ago

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Please help me with my science thank you!

Tatiana [17]

I cant see it maybe post it again?

4 0

3 years ago

What going on in each spot

KATRIN_1 [288]

nothing ksbsshshhzvsjajbsjshjsgdvdjhsbsj

3 0

3 years ago

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In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so

gulaghasi [49]

Answer: The moles of hydroxide ions present in the sample is 0.0008 moles

Explanation:

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 $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

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

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0

3 years ago