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tamaranim1 [39]
3 years ago
6

Vaccanation against the flu should not be required

Chemistry
1 answer:
sergiy2304 [10]3 years ago
4 0
Vaccination against the flu should not be required because:
- There have been recalls on vaccinations due to problems which will harm you more than the thing they are trying to prevent
- They are pure estimate since there are hundreds of forms of the flu, the percentage rate for success goes from about 50-80%, there have been years where they had 20% rates (aka nothing was prevented)
- Antibiotic resistance is growing and we are running out of vaccinations that work, lowering the amount of users would let the effects of a certain vaccination last longer
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Which mineral(s) might you be examining if you place some hydrochloric (hcl) acid on a sample and it fizzes?
makvit [3.9K]
The minerals that one might be examining if you place some hydrochloric acid on a sample and it fizzes are Calcite and Dolomite.
Calcite and dolomite are very similar minerals. Both have the same hardness, the same rhombohedral cleavage, and are found in identical geologic settings. The best way to tell one from the other is the acid test; A drop of 1 M HCl on calcite produces an instant , obvious Fizz; a drop on dolomite produces slow or no obvious bubbling. 
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2 years ago
What 3 sub types of covergent plate boundaries?
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Whose main job is to break down sugar to release energy that a plant cell can use?
Tasya [4]

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The Mitochondria

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3 years ago
How many orbitals are available in a “d” orbital configuration?
ratelena [41]

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8 0
3 years ago
A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th
Kisachek [45]

Answer:

pH=4.1

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1

Regards.

3 0
3 years ago
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