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leonid [27]
3 years ago
8

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 30.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?
Chemistry
1 answer:
mamaluj [8]3 years ago
4 0

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

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A hot air balloon starts with its temperature at 68.7°C and a pressure of 0.987 ATM and volume of 564L at what temperature in de
ICE Princess25 [194]

Answer:

54.7°C is the new temperature

Explanation:

We combine the Ideal Gases Law equation to solve this.

P . V = n. R. T

As moles the balloon does not change and R is a constant, we can think this relation between the two situations:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

T° is absolute temperature (T°C + 273)

68.7°C + 273 = 341.7K

(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂

1.63 atm.L/K = 532.5 atm.L / T₂

T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K

T° in C = T°K - 273 → 326.7K + 273 = 54.7°C

3 0
3 years ago
4.One rubber tire can generate about250,000 BTUS (1BTU=.0003KW ) when it is burned. The average American home consumes about 10,
jekas [21]

Answer:

The number of tires required to power ten homes for one year is approximately 2,730 tires

Explanation:

The given information are;

The energy generated by burning one rubber tire = 250,000 BTUs

1 BTU = 0.0003 kW

The amount of energy consumed by the average American home = 10,000 kWh

Therefore, the amount of energy generated by burning one tire in kW can be found as follows;

The energy generated by burning one rubber tire = 250,000 BTUs

1 BTU = 0.0003 kW

∴ 1 BTU × 250,000 = 250,000 BTUs = 0.0003 kW × 250,000 = 75 kW

We note that 250,000 BTUs is equivalent to 263764 kJ

1 kWh is equivalent to 3600 kJ

10,000 kWh = 10,000 × 3600 = 36,000,000 kJ

The energy requirement for ten homes  for one year = 10 × 36,000,000 kJ = 360,000,000 kJ

At 50% efficiency, the energy produced per tire = 0.5 × 263764 kJ = 131,882 kJ

The electrical energy produced per tire = 131,882 kJ/tire

The number of tires required to power ten homes for one year = 360,000,000 kJ/  131,882 kJ/tire = 2,729.71293  ≈2,730 tires

The number of tires required to power ten homes  ≈2,730 tires.

8 0
3 years ago
What is silicon used for? <br> matches<br> gunpowder <br> glass<br> steel can coating
Irina-Kira [14]

Answer:

steel can coating

Explanation:

7 0
3 years ago
Read 2 more answers
For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.
Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0
3 years ago
A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.
Neporo4naja [7]

Answer:

(a)  7.11 x 10⁻³⁷ m

(b)  1.11 x 10⁻³⁵ m

Explanation:

(a)  The de Broglie wavelength is given by the expression:

λ = h/p = h/mv

where h is plancks constant, p is momentum which is equal to mass times velocity.

We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.

v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s

m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg

λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m

(b) For this part we have to use the uncertainty principle associated with wave-matter:

ΔpΔx > = h/4π

mΔvΔx > = h/4π

Δx = h/ (4π m Δv )

Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.

Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )  

     = 0.045 m/s

Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )

     = 1.11 x 10⁻³⁵ m

This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.

5 0
3 years ago
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