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Answer:
1 mole
Explanation:
To get the answer to this, we need to write the correct chemical equation:
HCl + NaOH ------》NaCl + H2O
As seen in the balanced chemical equation above, one mole of hydrochloric acid will react with one mole of sodium hydroxide to yield one mole of sodium chloride and one mole of water
Boiling-point elevation is a colligative property.
That means, the the boiling-point elevation depends on the molar content (fraction) of solute.
The dependency is ΔTb = Kb*m
Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.
A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:
(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg = 50/molar mass
=> 200 / 50 = 4.
Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.
Or, what is the same, the solution with higher molality will have the higher boiling point.
Answer:
Catalyst is a substance that increases the rate of a chemical reaction but is chemically unchanged at the end of the reaction.
properties of catalyst :
1. A catalyst increases the speed of a reaction, and it also improves the yield of the intended product.
2. A catalyst actually takes part in the reaction even though it itself is not consumed or used up in the course of the reaction.
3. A catalyst makes the reaction faster by providing an alternative pathway with a lower activation energy.
4. A catalyst is reaction-specific. It may not be effective in another reaction even if the two reactions are of similar type.
5. In a reversible reaction, a catalyst accelerates both the forward and the reverse reactions. So, the inclusion of a catalyst does not alter the equilibrium constant of a reversible reaction.
Answer:
for the cellular respiration the human body needs 1.424 E21 molecules of glucose
Explanation:
mass glucose = 24 g
∴ Mw glucose = 180.156 g/mol
∴ 1 mol ≡ 6.022 E23 molecules
⇒ mol glucose = (24 g)×(mol/180.156 g) = 0.1332 mol glucose
⇒ molecules glucose = (0.1332 mol glucose)×(6.022 E23 molecules/mol)
⇒ molecules glucose = 1.424 E21 molecules