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3 years ago

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If a 12oz can of Soda contains 48.0g of sugar and you drink one can a day for one month (31 days), how many pounds (lb) of sugar

have you consumed from the soda alone? If 50% of this sugar is in the form of glucose (C6H12O6) how many grams of carbon have you consumed from glucose in the beverage over the 31 days?

1 answer:

marishachu [46]3 years ago

3 0

Answer:

yo dude on my questin can u answer my coment

Explanation:

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In which phase(s) of matter are the atoms closely packed but still able to slide past each other? gas liquid

lapo4ka [179]

The answer would be liquid.

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3 years ago

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How many moles of ammonium ions are in 125 mL of 1.40 M NH4NO3 solution? ________ moles (give answer with correct sig figs in un

Sholpan [36]

The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

Volume = 125 mL = 125 / 1000 = 0.125 L

Molarity = 1.40 M

<h3>Mole of NH₄NO₃ =? </h3>

Mole = Molarity x Volume

Mole of NH₄NO₃ = 1.40 × 0.125

<h3>Mole of NH₄NO₃ = 0.175 mole</h3>

Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:

NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)

From the balanced equation above,

1 mole of NH₄NO₃ contains 1 mole of NH₄⁺

Therefore,

0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺

Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

Learn more: brainly.com/question/25469095

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2 years ago

Write a procedure that you would follow to demonstrate the effect of increasingly higher concentrations of hydrochloric acid on

Hitman42 [59]

The reaction would produce bubbles of gas.

We can prepare 3-5 test tubes of acid with increasing concentrations. Then, we add antacid tablets to each and note the time taken for the tablet to dissolve and stop producing bubbles. The lesser the time taken, the greater the rate of reaction.

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3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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4 years ago

What ion does a base increase

cricket20 [7]

OH- is the ion that increases the concentration of a base

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