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3 years ago

SOMEONE PLS HELP ITS FOR A TEST!!

Chemistry

2 answers:

n200080 [17]3 years ago

8 0

4 is true and 5 is true

Damm [24]3 years ago

6 0

Explanation:

4=true

5. true

hope this helps. good luck with your test.

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When compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet,

kati45 [8]

Answer:

attached here is the diagram of the solution

Explanation:

4 0

3 years ago

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask

tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M$

0.18106 M is the molarity of the resulting solution.

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

$Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L$

0.823 Molar is the molarity of the solution.

6 0

2 years ago

Sucralfate (molecular weight: 2087 g/mol) is a major component of Carafate®, which is prescribed for the treatment of gastrointe

dybincka [34]

Answer:

Option b. 0.048 M

Explanation:

We have the molecular weight and the mass, from sulcralfate.

Let's convert the mass in g, to moles

1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.

Molarity is mol /L

Let's convert the volume of solution in L

10 mL . 1L/1000 mL = 0.01 L

4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L

3 0

3 years ago

Name the method that used to separated amino acids from fruit juice solution

Ket [755]

Using the method chromatography

8 0

3 years ago

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C

laila [671]

Answer:

First choice: 99 J/K

Explanation:

1) First law of thermodynamic (energy balance)

Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

2) Energy change of each substance:

General formula:

Heat released or absorbed = mass × Specific heat × change in temperature

density of water: you may take 0.997 g/ ml as an average density for the water.

mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

Specif heat of water: 1 cal / g°C

Heat released by the hot water:

Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

Heat absorbed by the cold water:

Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

Heat absorbed by the calorimeter

Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

4) Balance

Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

Convert to cal / K to Joule / K

1 cal = 4.18 Joule

23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0

3 years ago