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2 years ago

Where does the classification of species fall?

Chemistry

1 answer:

ivolga24 [154]2 years ago

5 0

Explanation:

develop a short message which you will deliver to the people in this community about application of chemistry in everyday life

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Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at

telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

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

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound Pressure at equilibrium

$H_2$ 61.8 atm

$I_2$ 46.5 atm

$HI$ 52.3 atm

Calculate the value of the equilibrium constant $K_p$ for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant $K_p$ for this reaction is, 0.952

Explanation :

The given chemical reaction :

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

The expression of $K_p$ for above reaction follows:

$K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}$

We are given:

$P_{H_2}=61.8atm$

$P_{I_2}=46.5atm$

$P_{HI}=52.3atm$

Putting values in above equation, we get:

$K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952$

Therefore, the value of equilibrium constant $K_p$ for this reaction is, 0.952

7 0

3 years ago

The closest star to our Sun is Proxima

lana [24]

answer:

not positive if this is correct just what i ould find on google

5 0

4 years ago

Which of the following choices has the ecological levels listed from largest to smallest?

wariber [46]

Answer:

the answer is something

Explanation:

3 0

3 years ago

Calculate number of moles of 4.4g of CO2 and 5.6L of NH3 . Pls help . Its urgent

kupik [55]

Moles of CO₂ = mass / molecular weight
Moles of CO₂ = 4.4 / (12 + 16 x 2)
Moles of CO₂ = 0.1 mol

Each mole of gas occupies 22.4 L at STP. Therefore,
Moles of NH₃ = 5.6 / 22.4
Moles of NH₃ = 0.25 mol

4 0

4 years ago

The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of ene

sesenic [268]

Answer:

$T_2=98.5^{\circ}$

Explanation:

Given that,

The specific heat of water is 4.184Jg°C

Mass, m = 600 g

Initial temperature, T₁ = 75.5°C

We need to find the final temperature. We know that heat absorbed is given by :

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}$

So, the final temperature is equal to $98.5^{\circ}$ .

3 0

3 years ago