A 23.6 ml of slat water has a mass of 26.47 grams. calculate its density.

Magnesium metal reacts with iodine gas at high temperatures to form magnesium iodide. what mass of mgi2 can be produced from the

sergiy2304 [10]

54.8 g of MgI2 can be produced. To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium Atomic weight of Iodine = 126.90447 Atomic weight of Magnesium = 24.305 Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394 Now determine how many moles of Iodine and Magnesium you have moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent. So figure out how many moles of magnesium will be consumed by the iodine 0.393997154 mole / 2 = 0.196998577 mole. This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2 0.196998577 mole * 278.11394 g/mole = 54.78805 g Round the result to the correct number of significant figures. 54.78805 g = 54.8 g

5 0

3 years ago

calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate

kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

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4 0

1 year ago

Need help on atoms and molecules

balu736 [363]

1-b

2-c

3-a

4-d

5-d

6-b

7-a

7 0

3 years ago

A gas mixture is composed of three different gases. The first gas in the mixture has a pressure of 0.35 atm. The second gas in t

riadik2000 [5.3K]

Answer:

1.20atm

Explanation:

Given parameters:

Partial pressure of gas 1 = 0.35atm

Partial pressure of gas 2 = 0.20atm

Partial pressure of gas 3 = 0.65atm

Unknown:

Total pressure of the gas mixture = ?

Solution:

To solve this problem, we need to recall and understand the Dalton's law of partial pressure.

Dalton's law of partial pressure states that "the total pressure of a mixture of gases is equal to the sum of the partial pressure of the constituent gases".

Total pressure =Pressure of gas(1 + 2 + 3)

The partial pressure is the pressure a gas would exert if it alone occupied the volume of the gas mixture.

Now we substitute;

Total pressure = (0.35 + 0.20 + 0.65)atm = 1.20atm

8 0

3 years ago

Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b.

Ilia_Sergeevich [38]

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) Trypsin

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) Chymotrypsin

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.

3 0

3 years ago