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Sloan [31]
2 years ago
15

Prior to a major outbreak of e. coli o157:h7 from hamburgers, the meat inspector's primary tools were:_____.

Chemistry
1 answer:
ser-zykov [4K]2 years ago
6 0

The principal equipment of the meat inspector before a significant E. coli O157:h7 outbreak linked to hamburgers was a sharp knife.

<h3>What is E. coli?</h3>

Escherichia coli (E. coli) is a type of bacterium that ordinarily dwells in both human and animal intestines. But some strains of E. coli, most notably E. coli O157:H7, can result in intestinal infection. After the toxin they generate, E. coli O157:H7 and other strains that cause intestinal illness are known as Shiga toxin-producing E. coli (STEC). Many patients who have an E. coli infection recover completely. However, it may result in serious, even fatal, problems. These problems are more likely to occur in older adults, pregnant women, young children, and people with compromised immune systems.

To know more about E. coli, visit:

brainly.com/question/1657681

#SPJ4

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What are variables?????
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A variable is an element, feature, or factor that is liable to vary or change.
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Which type of reaction involves the release of energy into the surroundings? Options: endothermic reactions
emmasim [6.3K]

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exothermic reaction

Explanation:

The definition of exothermic is:

describing a process that involves the release of energy into the surroundings by substances undergoing change.

Think of<em> ex</em>-othermic as energy exiting/ released and <em>en</em>-dothermic as energy entering.

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3 years ago
The job of the ______ is to hold all of the parts of a cell together
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Draw the Lewis structure for SiO2
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7 0
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g Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 0.96 g of methane
galben [10]

Answer: 2.64 g of carbon dioxide that could be produced by the chemical reaction.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} CH_4=\frac{0.96g}{16g/mol}=0.06moles

\text{Moles of} O_2=\frac{6.37}{32}=0.20moles

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

According to stoichiometry :

1 mole of CH_4 require = 2 moles of O_2

Thus 0.06 moles of CH_4 will require=\frac{2}{1}\times 0.06=0.12moles  of O_2

Thus CH_4 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

As 1 moles of CH_4 give = 1 mole of CO_2

Thus 0.06 moles of CH_4 give =\frac{1}{1}\times 0.06=0.06moles  of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.06moles\times 44g/mol=2.64g

Thus 2.64 g of carbon dioxide that could be produced by the chemical reaction.

7 0
4 years ago
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