Ask question

Ask question

All categories

3 years ago

14

A 0.0100 mol sample of Ca(OH)2 requires 50.00 mL of aqueous HCl for neutralization according to the reaction below. What is the

concentration of the HCl? Equation: Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)

1 answer:

harkovskaia [24]3 years ago

6 0

Answer:Cm=o,4 (Mol/l)

Explanation:

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

0,01 0,02

CM=n/V

You might be interested in

How do you separate sand and zinc?

taurus [48]

Answer:

By using magnet.

Explanation:

Put sand and zinc in the bowl.

Take a magnet an bring it close to the bowl.

The zinc from the sand will be attracted towards the magnet.

In this way it will be separated.

8 0

3 years ago

During aerobic respiration, electrons travel downhill in which sequence? A) food → citric acid cycle → ATP → NAD+

mina [271]

<span>During aerobic respiration, electrons travel downhill in which sequence?
</span><span>B) food → NADH → electron transport chain → oxygen</span>

7 0

3 years ago

Why are ionic compounds solids

enot [183]

Answer: The crystal structure of ionic compounds is strong and rigid. It takes a lot of energy to break all those ionic bonds. As a result, ionic compounds are solids with high melting and boiling points.

7 0

3 years ago

Read 2 more answers

What visible changes, in addition to precipitate formation, show that cations and anions have rearranged to form new products?

enyata [817]

Answer:

Precipitation reactions

Explanation:

6 0

3 years ago

HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into

dsp73

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.037 M

<u>Explanation:</u>

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

<u>Initial:</u> 1.0

<u>At eqllm:</u> 1.0-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$Kc=1.6\times 10^{-3}$

Putting values in above expression, we get:

$1.6\times 10^{-3}=\frac{x\times x}{(1.0-2x)^2}\\\\x=-0.043,0.037$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

Hence, the concentration of hydrogen gas at equilibrium is 0.037 M

6 0

3 years ago