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3 years ago

Why is Ca(OH)2 alkali?

1 answer:

4 0

Honestly I also have a question... Why is an elementary student learning high school level science. Back to the answer now; <span>An alkali is a basic hydroxide or </span>ionic salt of an alkali metal or alkaline earth metal element,<span> which is soluble in water. Hope this helped</span>

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Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

3 0

3 years ago

In the waves lab, which type of wave was shown on the oscilloscope screen?

Yanka [14]

Hi there!

I answered a question for you earlier and confused it with this question.
The answer is: B. Transverse

Happy learning!
~Brooke

4 0

3 years ago

The compound consists of 40.1% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative density with respect to hydrogen is -15. Wha

kap26 [50]

Answer:

https://socratic.org/answers/220339

this is the answer , my grandpa create this page

Explanation:

3 0

3 years ago

Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t

chubhunter [2.5K]

Answer:

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

Nair = 0.0206

PH20 = 0.19999

Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.

6 0

3 years ago

A sample contains 16. 75 g of the radioisotope U-236 and 50. 25 g of its daughter isotope, Th-232. How long did it take for deca

garri49 [273]

The time taken for the isotope to decay is 46 million years.

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 50.25 g

Amount remaining (N) = 16.75

Number of half-lives (n) =?

2ⁿ = N₀ / N

2ⁿ = 50.25 / 16.75

2ⁿ = 3

Take the log of both side

Log 2ⁿ = 3

nLog 2 = Log 3

Divide both side by log 2

n = Log 3 / Log 2

n = 2

Finally, we shall determine the time.

Half-life (t½) = 23 million years

Number of half-lives (n) = 2

Time (t) =?

t = n × t½

t = 2 × 23

t = 46 million years

Learn more about half-life: brainly.com/question/25927447

8 0

3 years ago

Read 2 more answers