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zvonat [6]
3 years ago
8

Which condition might develop if people eat contaminated raw or undercooked fish?

Chemistry
2 answers:
soldier1979 [14.2K]3 years ago
7 0
The correct answer is Anosamiasis
White raven [17]3 years ago
5 0
O B. Anisakiasis, you get it from eating undercooked or raw fish
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Which type of reactions form salts?
Marysya12 [62]

Answer:

Neutralization reactions

Explanation:

A neutralization reaction is a reaction between an acid and a base. Products of this type of reaction is water and a salt. The pH of the salt product would depend on how strong or weak the base and acid would be when they react with each other. Although the characteristics of bases and acids are practically polar opposites, when combined, they cancel each other our producing a neutralized product.

6 0
3 years ago
Determine the oxidation number for nitrogenin
iris [78.8K]

Answer:

1(a) N = 3

(b) N = 0

(c) N = 5

(d) N = -2

(2) Molecular formula for benzene is C6H6

Explanation:

1(a) N02 1-

N + (2×-2) = -1

N-4 = -1

N = -1+4 = 3

(b) N2

2(N) = 0

N = 0/2 = 0

(c) NO2Cl

N + ( 2×-2) + (-1) = 0

N - 4 - 1 = 0

N - 5 = 0

N = 0+5 = 5

(d) N2H4

2(N) + (4×1) = 0

2N + 4 = 0

2N = 0 - 4 = -4

N = -4/2 = -2

(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole

Therefore, molecular formula for benzene is C6H6

7 0
3 years ago
How likely is it that a scientific theory is correct?
yan [13]
It all depends what theory it is most are supported by really good evidence but they just don't have all the evidence so it can't be proven a fact at that time

5 0
3 years ago
I am so bored someone talk
kolezko [41]

Answer:

hello.

Explanation:

you wanted to talk? well here I am

5 0
2 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
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