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anygoal [31]
2 years ago
6

How would an apple, potato, and onion all taste the same?

Chemistry
2 answers:
castortr0y [4]2 years ago
6 0

Answer:

A

Explanation:

Says it on the side of meh screen :P

(Ez question in lol)

muminat2 years ago
6 0

Answer:

A. Plugging your nose.

Explanation:

It is frequently quoted that upwards of 80% of our taste is made up by smell. So if you plug your nose and cover your eyes, the taste between an apple and onion should be indistinguishable. All potato chip flavors taste the same.

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Points + Brainalist
Lorico [155]

Answer:

I think A it looks to be the answer

Explanation:

sorry if wrong

4 0
2 years ago
A buffer is made by dissolving H3PO4 and NaH2PO4 in water.
xeze [42]

Answer:

(a) H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)

(b) OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)

Explanation:

The equation for your buffer equilibrium is:

H₃PO₄(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq)+ H₂PO₄⁻(aq)

(a) Adding H₃O⁺

The hydronium ions react with the basic dihydrogen phosphate ions.

H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)

(b) Adding OH⁻

The OH⁻ ions react with the more acidic hydronium ions.

OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)

4 0
3 years ago
Read 2 more answers
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
what mass of small stones would be needed to make the strongest concrete? give a reason for your awnser.
VikaD [51]

Answer:

For a good consert mix aggregate needed to the clean hard strong partical free of absorb chemicals or coating clay and other fine materials

3 0
3 years ago
PLEASEEE HELPP I BEGG FOR HELPPP
Basile [38]

Answer:

An element that is oxidized is a reducing agent, because the element loses electrons, and an element that is reduced is an oxidizing agent, because the element gains electrons.

7 0
3 years ago
Read 2 more answers
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