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Zarrin [17]
3 years ago
7

A 11.630 g milk chocolate bar is found to contain 7.815 g of sugar.How many milligrams of sugar does the milk chocolate bar cont

ain?
Chemistry
2 answers:
tigry1 [53]3 years ago
6 0

Answer: 7815 mg

Explanation:

Mass is the amount of matter contained in a substance.

It can be measured in units of kilograms (kg), grams (g), milli grams (mg) etc;

These units are interconvertible.

As it is given,  11.630 g milk chocolate bar is found to contain 7.815 g of sugar

milligrams of sugar= ?

1 g= 1000 mg

7.815g=\frac{1000}{1}\times 7.815=7815mg

Thus milk chocolate bar contain 7815 mg of sugar.

zzz [600]3 years ago
4 0
3.815 that's the awnser......................................




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Kinetic energy <br><br>definition:<br><br><br>sentence: ​
Irina18 [472]

Answer:

Kinetic energy , is the energy of an object possessed due to it's motion

Explanation:

Kinetic energy is the energy of mass in motion. The kinetic energy of an object is the energy it has because of its motion.

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7 0
3 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

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3 years ago
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Answer:

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The structures of both acetone and propanal are shown below:

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