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3 years ago

The stomach lining is made up of deep muscular grooves. How do you think these structures help the stomach to break down food

1 answer:

6 0

The curves absorb the food and sort of squish it and make it easier to digest then the food goes into the stomach acid and kind of burns and become and little softer<span> or in other words by allowing the stomach to stretch in order to accommodate large meals and get grip and move food during the digestion process. </span>

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How many grams of c2h5oh must be burned to raise the temperature of 400.0 ml of water from 20.0 ∘c to 100.0 ∘c? (the specific he

Lana71 [14]

From tables, the heat of combustion of ethanol (C₂H₅OH) is 29.7 kJ/g.

Given:
V = 400 mL of water
ΔT = 100 - 20 = 80°C, temperature rise
c = 1.00 cal/(g-°C), specific heat of water = 4.184 J/(g-°C)

Because the density of water is approximately 1.0 g/mL. the mass of water is
m = 400 g.

Let x = grams of ethanol that should be burned to make the water rise
by 80 °C.
Then
(29.7 x 10³ J/g)*(x g) = (400 g)*(4.184 J/g-°C))*(80 °C)
29.7 x 10³x = 1.339 x 10⁵
x = 4.5 g

Answer: 4.5 grams

7 0

4 years ago

If 200 grams of hydrogen gas react with 150 grams of oxygen gas, then how many grams of water will be produced

Allushta [10]

2H2 + O2 -> 2H20 is the balanced equation for the reaction of Hydrogen with Oxygen to form water so If you have 32g of O2 this is a simple Dimensional analysis problem 32g O2 x 36.03056g H20/31.9988g O2 this way the O2 cancels out and you are left with just the H2O so your raw answer would be 36.0319112, then if your instructor requires a significant figure answer that would be to 2 significant figures the information you were given 32g O2, so as above 36g or Water are produced. Just a different way to view and solve the problem with the balanced equation so you can see the way everything relates to everything else. the molar masses of O2 and H2O are simply found my adding up 2 Oxygens 15.9994g x2 = 31.9988g and H2O = 2(1.00794) + 15.9994 = 18.01258 but you then have to multiply that by 2 because the reaction states you get 2 mols in the reaction so that is where I came up with the 36.03056g for the solution. Hope this helps. Dimensional Analysis is and can be one of the best ways to solve these problems, because not always are you going to be dealing with 1:2 ratios.

6 0

4 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg o

aev [14]

n - number of mole

M(CO2) = 12.0 +16.0*2 = 44.0 mg/mmol

M(H2O) = 2*1.0 +16.0= 18.0 mg/mmol

1) 14.242 mg of CO2*1 mmol/44.0g =0.32368 mmol CO2

0.32368 mmol CO2 has 0.32368 mmol C

2) 0.32368 mmol C*12 mg/1 mmol= 3.88418 mg C

3) 4.083 mg of H2O * 1 mmol/18 mg = 0.226833 mmol H2O

0.226833 mmol H2O has 2*0.226833 mmol H = 0.453666 mmol H

4) 0.453666 mmol H*1 mg/1mmol = 0.453666 mg H

5) Mass of N = 5.250 mg - 3.88418 mg (C) - 0.453666 mg (H)≈ 0.912 mg N

0.912 mg N * 1 mmol/14.0 mg = 0.0652 mmol N

6) n(C) : n(H) : n(N) = 0.32368 mmol C : 0.453666 mmol H : 0.0652 mmol N

0.0652 mmol N is a smallest value.

n(C) : n(H) : n(N) = (0.32368 mmol/0.0652 mmol) C : (0.453666 mmol H/0.0652 mmol) : (0.0652mmol/0.0652 mmol)N = 5 : 7 : 1

n(C) : n(H) : n(N) = 5 : 7 : 1

Empirical formula : C5H7N.

8 0

3 years ago

What are the four season of the northern hemisphere .

Ksju [112]

Answer:

spring, summer, fall, and winter

5 0

3 years ago

Read 2 more answers

In the titration of 50.0 mL of HCl of unknown concentration, the phenolphthalein indicator present in the colorless solution tur

KIM [24]

Answer:

The molarity of HCl is 0.138 M

Explanation:

The titration reaction is as follows:

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:

At the endpoint:

moles Ca(OH)₂ = moles HCl / 2

Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:

mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂

mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂

The number of moles of HCl will be:

mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl

This number of moles was present in 50.0 ml, then, in 1000 ml:

mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol

Then:

Molarity HCl = 0.138 M

7 0

4 years ago