Answer:
6.022 ×10(index 23) / 7.5 = 0.8293 ×10(index 23)
Explanation:
molar mass of C = 12gmol
therefore in 12g of C there is one mole or an amount of 6.022 ×10(index 23)
∴12g/6.02210(index 23) ×1.6g
Answer:
1/3
Explanation:
Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.
Pyruvate is converted to acetyl-CoA in the reaction given below:
Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂
1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.
Also,
2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).
Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.
Thus pyruvate produces total 3 molecules of CO₂ and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)
Thus,
<u>Fraction = 2/6 = 1/3</u>
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1
number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂
we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation
101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL
When sodium carbonate is dissolved in water, the equation is 
.
When carbon dioxide is placed in water, aqueous carbon dioxide is formed: 
<h3>Dissolution of compounds in water</h3>
Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.
Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.
While in solution, sodium carbonate dissociates into its component ions according to the following equation:
Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.
Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as:
More on the dissolution of substances can be found here: brainly.com/question/28580758
#SPJ1
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For N₂,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference
0.00 (Non Polar Covalent)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic)
For CO₂,
E.N of Oxygen = 3.44
E.N of Carbon = 2.55
________
E.N Difference 0.89 (Polar Covalent)
