What volume of 0.585 m ca(oh)2 would be needed to neutralize 15.8 l of 1.51 m hcl?

1 answer:

8 0

When the balanced reaction equation is:

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2

2:1

∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) * (1L ca(OH)2/0.585 mol Ca(OH)2

= 20.4 L

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Need asap thank you

cupoosta [38]

Answer:

c.boron-11

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The atomic mass of boron is 10.81 u.

And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.

To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:

(10u)(x)+(11u)(1−x)100%=10.81u

Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.

Solving for x we get:

11u−ux=10.81u

0.19u=ux

x=0.19

1−x=0.81

And thus the abundance of boron-11 is roughly 81%.

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The combustion of palmitic acid is represented by the chemical equation: C16H32O2(s) + 23O2(g) → 16 CO2(g) + 16 H2O(l) The magni

solniwko [45]

Answer:

The correct option is C

Explanation:

From the question we are told that

The reaction is

$C_{16}H_{32}O_2(g) + 23O_2(g) \to 16 CO_2(g) + 16 H_2O(l)$

Generally $\Delta H = \Delta U + \Delta N* RT$

Here $\Delta H$ is the change in enthalpy

$\Delta U$ is the change in the internal energy

$\Delta N$ is the difference between that number of moles of product and the number of moles of reactant

Looking at the reaction we can discover that the elements that was consumed and the element that was formed is $O_2$ and $CO_2$ and this are both gases so the change would occur in the number of moles