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3 years ago

Can You Tell? Pepsinogen is released in the stomach. Why can't it digest stomach walls?

Chemistry

1 answer:

miv72 [106K]3 years ago

5 0

Answer:

Specific cells within the gastric lining, known as chief cells, release pepsin in an inactive form, or zymogen form, called pepsinogen. By doing so, the stomach prevents the auto-digestion of protective proteins in the lining of the digestive tract.

Explanation:

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what is the proper scientific notation for 0.000476? a.0.476 × 10-3 b.4.76 × 10-4 c.47.6 × 10-5 d.476 × 10-6

-BARSIC- [3]

B. 4.75 × 10-4 is the answer

6 0

3 years ago

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

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3 years ago

The enthalpy of formation of xef2(g) is –108 kj mol–1 and the bond dissociation enthalpy of the f–f bond is 155 kj mol–1 . what

S_A_V [24]

Xe +f2 →Xef2
ΔXe = ΣB.P reactants - Σ B.d products
-108k.s/ mol = B. D f₂ - 2 B.D xe-f
-108 k.s/mol =155 k.s/mol - 2B.Dxe-f
263kJ/mol/2 = B. D xe-f
B.D xef = 131.5 kJ/mol
132 kJ/mol

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3 years ago

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Oduvanchick [21]

Answer:

Answer is B.

Explanation:

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3 years ago

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Assoli18 [71]

Hmm well thinking of how amperes are generated it should be true

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3 years ago

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