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Viefleur [7K]
3 years ago
14

How many moles are in 9.8 x 1024 atoms of calcium?

Chemistry
1 answer:
Ratling [72]3 years ago
4 0

Explanation:

no of moles = no of atoms ÷ avogadro's number

= (9.8×10^24) ÷ (6.02×10^23)

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Sometimes described as a "chemical brother" of gaba (gamma-aminobutyric acid), which amino acid neurotransmitter acts in an exci
poizon [28]

excitatory amino acids are the amino acids helps in transformation of neurotransmitters or it helps in transmission of synapsis rapidly in brain of mammal. EAA known to be neurotransmitters for Central nervous system.

Excitatory amino acids count may vary from 50’s to 100’s. They are mostly composed of non-protein- amino acids obtained from algae or fungi.

The possible EAAs are Glutamate (Glu) and Aspartate which act as excitatory neurotransmitters in the brain. They get released from neurons where they induce excitation via metabotropic Glu receptors.  

Both glutamate and aspartate having excitatory effect on neurotransmission whereas Gama-amino butyric acid having inhibitory effect on neurotransmission.

Thus, the statement ‘neurotransmitters are chemical brother of gaba’ is indicating the complementary effect of each other.


4 0
3 years ago
Water is believed to be one of the best fire extinguishers then why it is so that fire in oil spills over ocean/sea is difficult
Eddi Din [679]

Answer:

D

Explanation:

Trust me

6 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
Given the equation representing a solution equilibrium: BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq)
goldenfox [79]

Answer:

The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases

Explanation:

Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.

In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;

BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.

This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.

3 0
3 years ago
Determine the boiling point of a 3.70 m solution of phenol in benzene. Benzene has a boiling point of 80.1°C and a boiling point
xeze [42]

Answer: The boiling point of a 3.70 m solution of phenol in benzene is 89.5^0C

Explanation:

Elevation in boiling point:

\Delta T_b=i\times k_b\times m

where,

\Delta T_b = change in boiling point

i= vant hoff factor = 1 (for benzene which is a non electrolyte )

k_b = boiling point constant = 2.53^0C/kgmol

m = molality = 3.70

T_{solution}-T_{solvent}=i\times k_b\times m

T_{solution}-80.1^0C=1\times 2.53\times 3.70

T_{solution}=89.5^0C

Thus  the boiling point of a 3.70 m solution of phenol in benzene is 89.5^0C

8 0
3 years ago
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