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patriot [66]
2 years ago
8

Explain why the grams of the nutrient molecules in a food do not add up to the total gram weight of the food.

Chemistry
2 answers:
nalin [4]2 years ago
6 0

Answer:

Explanation:

The food contains the large proportion of water and added insoluble fibers which also contains essential nutrients does not add up much to the total gram weight of the food.

leva [86]2 years ago
5 0

When we talk about the gram weight of any food item, it may or may not include the gram weight of its different nutrient molecules. It may have different reasons to it. One of the reasons can be because the nutrient molecule might contain a large percentage or proportion of water. Also, the nutrient molecules might be insoluble or indigestible fiber so it does not add up to the total gram weight of food.

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Can someone help me with this please
never [62]

Answer: B. 1:2

Explanation: Beryllium and chlorine forms a binary ionic compound. Ionic compound is formed when a metal loses its electrons to a receiving non metal. Beryllium (metal) has two valence electrons while chlorine (nonmetal) has seven valence electrons, and so a beryllium atom has to give out its two valence electrons to attain a duplet stable structure while a chlorine atom will gain one electron to attain its stable octet structure. In the reaction between beryllium and chlorine, two atoms of chlorine have to accept the two electrons from one beryllium atom to attain their stable octet structure.

The formula of the compound formed is BeCl2.

3 0
3 years ago
Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g H2 to produce water.
kupik [55]
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,

                    2H2 + O2 --> 2H2O

The masses of each of the reactants are calculated below.

          2H2 = 4(1.01 g) = 4.04 g
          O2 = 2(16 g) = 32 g

Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
        (1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2

Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.


<em>Answer: 1.22 g of oxygen</em>
4 0
3 years ago
How many of the following are found in 15.0 kmol of xylene (C8H10)? (a) kg C8H10; (b) mol C8H10; (c) lb-mole C8H10; (d) mol (g-a
77julia77 [94]

Answer:

a) 1592.4 kg C8H10

b) 15*10³ mol C8H10

c) 33.1 lb-mole

d) 1.2 *10^5 mol C

e)  1.5 * 10^5 mol H

f)  1.44 * 10^6 grams C

g) 1.52 * 10^5 grams H

h) 9.03*10^27 molecules C8H10

Explanation:

Step 1: Data given

Number of moles C8H10 = 15.0 kmol =15000 moles

Molar mass of C8H10 = 106.16 g/mol

Step 2: Calculate mass C8H10

Mass C8H10 = moles C8H10 * molar mass C8H10

Mass C8H10 = 15000 * 106.16 g/mol

Mass C8H10 = 1592400 grams = 1592.4 kg

Step 3: Calculate moles C8H10

15.0 kmol = 15*10³ mol C8H10

Step 4: Calculate lb-mol

15.0 kmol = 33.1 lb-mole

Step 5: Calculate moles of C

For 1 mol C8H10 we have 8 moles of C

For 15*10³ mol C8H10 we have 8* 15*10³  =1.2 *10^5 mol C

Step 6: Calculate moles H

For 1 mol C8H10 we have 10 moles of H

For 15*10³ mol C8H10 we have 10* 15*10³  = 1.5 * 10^5 mol H

Step 7: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 1.2 *10^5 mol C * 12 g/mol

Mass C = 1.44 * 10^6 grams

Step 8: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 1.5 *10^5 mol H * 1.01 g/mol

Mass C = 1.52 * 10^5 grams

Step 9: Calculate molecules of C8H10

Number of molecules = number of moles * number of Avogadro

Number of molecules =  15*10³ mol C8H10 * 6.022*10^23

Number of molecules = 9.03*10^27 molecules

8 0
3 years ago
How much heat is required to raise the temperature of 50.0g of water 13°C<br>​
N76 [4]

Answer:one gram by 1oC

Explanation: you will need to know the value of water's specific heat

3 0
3 years ago
what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?
LuckyWell [14K]
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.
7 0
3 years ago
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