How do you determine the type of ion charge any element will form based on its number of valence electrons?

How would you make 5mls of a solution that is 2.0% lactose and 0.1M SPG buffer from separate stock solutions that are 6% lactose

Aliun [14]

Answer:

Check the explanation section.

Explanation:

The following steps should be followed orderly.

STEP ONE:

Use the dilution equation in the calculation of the volume for the stock solution. That is, C1 × V1 = C2 × V2.

Where C1 and C2 are the concentration of the stock solution and the diluted solution.

STEP TWO:

Put 6% of lactose and make sure to dilute it in order to make 2.0% lactose

and put it in Beaker A. Also, make sure to dilute the 1M to 0.1M SPG buffer in Beaker B.

STEP TWO:

Now, from beaker A containing 2% lactose, measure and remove 5.0 mL from it. Also, measure and remove 5.0 mL from beaker B containing 0.1M SPG.

So, in STEP TWO above we won't know how much water we need to use for dilution, thus, there is the need to make use of STEP ONE.

Therefore, from STEP ONE ABOVE, we have the dilution equation given as;

C1 × V1 = C2 × V2.

Hence, 6 × V1 = 2 × 5. Therefore, the volume needed from the stock solution, V1 = (2 × 5)/ 6 = 1.6 mL.

STEP THREE:

Now measure out 1.6 mL from the stock solution, that is 6% lactose and add it to 5mL of the diluted solution of 2% in beaker A into another container, say beaker C and add H2O to form SOLUTION X.

STEP FOUR:

Using the dilution equation again, Determine the the volume that is needed from 1M SPG.

C1 × V1 = C2 × V2.

V1 = ( 0.1 × 5)/ 1 = 0.5mL.

STEP FIVE:

measure 0.5mL out from the 1M SPG and 5 mL out of 0.1M SPG buffer and add water to it to form SOLUTION Y.

STEP SIX:

Now, mix solution X and solution Y together and take the required 5ml

7 0

3 years ago

Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.0000524 mol K I O 3 were titrated with an u

11111nata11111 [884]

Answer: The amount of sodium thiosulfate required is $2.62\times 10^{-5}$ moles

Explanation:

Moles of $KIO_3$ solution given = 0.0000524 moles

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.0000524 moles of potassium iodate will react with = $\frac{1}{2}\times 0.0000524=0.0000262mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $2.62\times 10^{-5}$ moles

4 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

As ice melts from a solid to a liquid, A) entropy increase

Dovator [93]

That answer is false

4 0

3 years ago

The Tin Pan Alley era lasted from 1950 to 1967. a. true b. false

fenix001 [56]

I think false is the answer .if wrong correct meeee

7 0

3 years ago