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Masteriza [31]
2 years ago
14

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

into the wound. If 20.0 g is reacted with 40.0 g , what mass of silver sulfadiazine, , can be produced, assuming 100% yield?
Chemistry
1 answer:
Katen [24]2 years ago
6 0

Answer:

We will produce 57.1 grams of silver sulfadiazine

Explanation:

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If 20.0 g of Ag2O is reacted with 40.0 of C10H10N4SO2, what mass of silver sulfadiazine (AgC10H9N4SO2) can be produced assuming 100% yield?

Step 1: Data given

Mass of Ag2O = 20.0 grams

Molar mass of Ag2O = 231.735 g/mol

Mass of C10H10N4SO2 = 40.0 grams

Molar mass of C10H10N4SO2 = 250.277 g/mol

Step 2: The balanced equation

Ag2O + 2C10H10N4SO2 → 2AgC10H9N4SO2 + H2O

Step 3: Calculate moles Ag2O

Moles Ag2O = mass Ag2O / molar mass Ag2O

Moles Ag2O = 20.0 grams / 231.735 g/mol

Moles Ag2O = 0.0863 moles

Step 4: Calculate moles C10H10N4SO2

Moles C10H10N4SO2 = 40.0 grams / 250.277 g/mol

Moles C10H10N4SO2 = 0.160 moles

Step 5: Calculate the limiting reactant

For 1 mol Ag2O we need 2 moles 2C10H10N4SO2 to produce 2 moles AgC10H9N4SO2 and 1 mol H2O

C10H10N4SO2 is the limiting reactant. There will react 0.160 moles.

Ag2O is in excess. There will react 0.160/ 2 = 0.080 moles

There will remain 0.0863 - 0.080 = 0.0063 moles Ag2O

Step 5: Calculate moles AgC10H9N4SO2

For 1 mol Ag2O we need 2 moles 2C10H10N4SO2 to produce 2 moles AgC10H9N4SO2 and 1 mol H2O

For 0.160 moles C10H10N4SO2 we'll have 0.160 moles AgC10H9N4SO2

Step 6: Calculate mass AgC10H9N4SO2

Mass AgC10H9N4SO2 = moles * molar mass

Mass AgC10H9N4SO2 = 0.160 moles * 357.137 g/mol

Mass AgC10H9N4SO2 = 57.1 grams

We will produce 57.1 grams of silver sulfadiazine

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Based on the following molecular weight data of polypropylene, determine the degree of polymerization Molecular Weight Range (g/
Lady bird [3.3K]

Answer:

785

Explanation:

Molecular. X. W

Weight

8000-16000 0.05 0.03

16000-24000. 0.017. 0.08

24000-32000. 0.22. 0.18

32000-40000. 0.25. 0.35

40000-48000. 0.22. 0.27

48000-56000. 0.09. 0.09

Mean weight X*M. W*M

12000. 600. 240

20000. 3200. 2000

28000. 6720. 5600

36000. 10080. 10800

44000. 8800. 11880

52000. 3640 3640

Total=33040g\mol 36240

Note before repeat molecular weight m= 3*12.01+6*1.008=

42.08g/mol

Degree of polymerization= total W*M/w=33040/42.08 =785

8 0
3 years ago
Convert 0.95 kilograms into centigrams. (1 g = 100 cg; 1 kg = 1000 g)
svet-max [94.6K]

Answer:

95,000 centigrams

Explanation:

There is 1000 CG in 0.01 kilograms

so you do 1000*95 which equals 95,000 centigrams.

6 0
3 years ago
3 NO2(g) + H2O(ℓ) −→
gulaghasi [49]
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L
7 0
3 years ago
What is the solubility product (or ion-product), Ksp, expression at equilibrium for calcium phosphate, Ca3(PO4)2?
Gnoma [55]

Answer : The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2

Explanation :

Solubility product : It is defined as the product of the concentration of the ions  that present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only.

The solubility product constant is represented as, K__{sp}.

The dissociation of calcium phosphate is written as:

Ca_3(PO_4)_2\rightleftharpoons 3Ca^{2+}+2PO_4^{3-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2

4 0
3 years ago
A 55-mL solution of H2SO4 is completely neutralized by 46 mL of 1.0M NaOH. What is the concentration of the H2SO4∆H2SO4(aq) +2Na
slega [8]
H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O

v(NaOH)=46 ml=0.046 l
c(NaOH)=1.0 mol/l
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n(NaOH)=v(NaOH)*c(NaOH)

n(H₂SO₄)=0.5n(NaOH)

c(H₂SO₄)=n(H₂SO₄)/v(H₂SO₄)=0.5*v(NaOH)*c(NaOH)/v(H₂SO₄)

c(H₂SO₄)=0.5*0.046*1.0/0.055=0.418 mol/l

The concentration of the H₂SO₄ is 0.418M.

4 0
3 years ago
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