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2 years ago

If you were a scientist investigating the effects of antioxidants in foods, what would you test and why?

Chemistry

1 answer:

lina2011 [118]2 years ago

5 0

Answer:

One possible reason why many studies on antioxidant supplements do not show a health benefit is because antioxidants tend to work best in combination with other nutrients, plant chemicals, and even other antioxidants.

Explanation:

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calculate the volume that will be occupied by 350 ml of oxygen measured at 720 mm hg, when the pressure changed to 630 mm hg

likoan [24]

Answer:

406.45mL

Explanation:

The following data were obtained from the question:

V1 = 350mL

P1 = 720mmHg

P2 = 630mmHg

V2 =?

The new volume can be obtain as follows:

P1V1 = P2V2

720 x 350 = 620 x v2

Divide both side by 620

V2 = (720 x 350) /620

V2 = 406.45mL

The new volume of the gas is 406.45mL

8 0

3 years ago

What is the condensed structural formula of the product of the reaction of 2,7-dimethyl-4-octene with hydrogen and a metal catal

gavmur [86]

The condensed structural formula of the product of the reaction of 2,7-dimethyl-4-octene with hydrogen and metal catalyst.

Ch3 CH(CH3) CH2 CH2 CH2 CH2 CH(CH3) CH3

Equation is as follows

CH3 CH(CH3) CH2 C=C CH2 CH(CH3) CH3 + H2→

CH3 CH(CH3)CH2 CH2 CH2 CH2 CH(CH3) CH3

metal catalyst example is nickel and the name of structure formed is
2,7- dimethyl octane

6 0

3 years ago

How is the way a mixture is combined different from how a compound is combined?

artcher [175]

Mixture is composed of molecules of different types. A compound can only be separated through chemical means. While a compound is a pure substance that contains 2 or more elements chemically combined together while a mixture are formed when two substances are added together without chemical bonds being formed.

7 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

Help me ASAP PLS

n200080 [17]

Answer:

18 g

Explanation:

We'll begin by converting 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:

Volume = 0.5 L

Molarity = 0.2 M

Mole of C₆H₁₂O₆ =?

Molarity = mole / Volume

0.2 = Mole of C₆H₁₂O₆ / 0.5

Cross multiply

Mole of C₆H₁₂O₆ = 0.2 × 0.5

Mole of C₆H₁₂O₆ = 0.1 mole

Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.1 mole

Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mass = mole × molar mass

Mass of C₆H₁₂O₆ = 0.1 × 180

Mass of C₆H₁₂O₆ = 18 g

Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.

6 0

3 years ago