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Ber [7]
3 years ago
15

Fats help your brain and nervous system develop correctly. True False

Chemistry
2 answers:
AURORKA [14]3 years ago
7 0
True


Yes that’s correct
Fantom [35]3 years ago
6 0

Answer:

Yes

Explanation:

the amino acids in proteins and the fatty acids in fat replace the amino and fatty acids in your brain to help you function and develop correctly.

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I need help please ​
Alex17521 [72]

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

4 0
3 years ago
What mass of H2O is produced by the combustion of 1.00 mol of CH4?
kkurt [141]
CH4 : H2O
1 : 2

number of moles of H2O = 1.00 x 2
number of moles of H2O = 2.00mol

mass = number of moles x molar mass

mass of H2O = 2.00 x (1 + 1 + 16)
mass of H2O = 36g
8 0
2 years ago
The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

6 0
3 years ago
Sha-vzwq-aby<br><br> Ladkiyoooooo aaa jaooo♡♥︎♡<br><br> Jaldiiiii karo na yaaaar
leva [86]

Answer:

kjajjahahayq :/

Explanation:

a sbywsbgv usnwbhx hg xw nx hb gs

5 0
2 years ago
Read 2 more answers
Which statement below matches the correct response with the proper reasoning when comparing the volatility of CH2Cl2 with CH2Br2
Nutka1998 [239]

Answer:

b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂

Explanation:

CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE

Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.

There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.

I hope it helps!

5 0
3 years ago
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