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3 years ago

Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.

Chemistry

1 answer:

Nataly_w [17]3 years ago

6 0

Answer:

$m_{sugar}=23.7g\ sugar$

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

$8oz*\frac{29.5735mL}{1oz} =236.6mL$

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

$m_{sugar}=236.6mL*\frac{10g\ sugar}{100mL}\\ \\m_{sugar}=23.7g\ sugar$

Best regards.

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Answer:

Option C. The same number of energy levels.

Explanation:

From the diagram given above, element (i) belong to group 2 while element (ii) belong to group 6.

Also, both element i and ii belong to the same period (i.e period 4). This simply means that both element i and ii have the same number of energy levels.

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3 years ago

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

$\frac{44}{56}$

= 0.785 g of CO2

So,

14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g of CaO produce =

$\frac{100}{56}$

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

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$\frac{19.4}{25.71}\times 100$

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3 years ago

How many moles of methane are in 7.31x10^25 molecules?

lana [24]

Answer:

121.37 moles.

Explanation:

Always one mole of any molecule contains Avogadro number of molecules.

So one mole of methane contains 6.023*10^23 molecules.

Now we need number of moles of methane in 7.31*10^25 molecules.

This is just cross multiplication stuff.

6.023*10^23 molecules —————- 1 mole

7.31*10^25 molecules ——————- ?

= (7.31*10^25) / (6.023*10^23)

= 121.37 moles.

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3 years ago

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2 years ago

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Harlamova29_29 [7]

Answer:

45.77 × 10²³ atoms

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Given data:

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The given problem will solve by using Avogadro number.

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3 years ago

Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.​

Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.